 So trigonometric equations and in equations trigonometric equations and in equations. Okay, so we are not, we are not only going to talk about equations, but we are also going to talk about in equations. As you only know in equations are nothing but inequalities involving variables. Right, so how to solve trigonometric equations and in equations is what we are going to learn in this chapter. Okay. So, in order to explain this concept, in order to start with this topic, I would ask you a very simple question. Okay, let us say there is a variable X and sign of that variable X is half. Okay, just a simple question. Sign X is equal to half. X is a variable. Now, what do you think are the possible values of X? If possible, please give me a response and radiance. Okay, don't give it in degrees. So what are the possible values of X that satisfy this equation? By the way, are there just one or two values or there are many values? What do you think? Let's answer that question first. Are there one or two values or there could be many values? There are many values, right? Why, why there are many values? Why do you think this equation will have many values? Right, now please understand the trigonometric equations are different from the equations that you normally see in algebra or maybe let's say if you have solved equations involving polynomials let's say transcendental functions earlier, any equation that you have seen before would have limited number of solutions. Of course, unless until it's an identity, identity is having infinitely many solutions. In fact, it is true for all the values of the variable. So trigonometric equations are slightly different in the sense that they are periodic functions. Okay, so by the virtue of the fact that all trigonometric functions are periodic. Okay, periodic means they repeat themselves after a certain fixed change in the value of the variable, right? So sin X is one function which is periodic with a period of 2 pi, correct? Cos X is also periodic with 2 pi. In fact, C kicks and C kicks they are also periodic with 2 pi and tan X and cortex they are periodic with pi each. Okay, if you want, I can write them down for you. So let me just list down function and their period. Period is something which you will be talking in more details in class 12 under the function chapter. So all the six trigonometric functions, whether it is sine or cos or coseak or seek or tan or cot, they have all been known to be periodic with a period as mentioned 2 pi, 2 pi, 2 pi, 2 pi, pi and pi. That means let's say tan of a particular value is K. Even if you change that value to, you know, that value plus pi or minus pi, the answer of that will still be K. Okay, so that is the meaning of periodicity. So periodicity, everybody knows that even if you change your X by X plus T, it is going to give you the same value of the function back. Okay, so this T here is called the period of the function. We will talk more in details of it. Okay, when we go to class 12 under the functions topic. So the problem with this trigonometric equation is because they are periodic, there will be infinitely many solutions possible. But nevertheless, it doesn't make them identity. Don't worry. Identity is where it is going to be satisfied for all the values of the variable provided that function is a defined one. But despite not being an identity also, there would be infinitely many answers possible to such an equation. And that is because these functions are known to be periodic. So let me show this scenario on a graph. So Sinex graph, if you all recall, looks like this. It's a sinusoidal graph. I'm just drawing a few branches of it. Okay, so this is your sinusoidal graph. Okay, so this goes on and on on both directions. This is the graph of Sinex versus X. Okay, now when you're trying to solve Sinex equal to half, basically you're trying to see. Basically you're trying to see what are the what are the possible values of X for which this line, which is your y equal to half line. Okay. Into six, the Sinex graph right solving an equation is what solving an equation is trying to see solving this is basically trying to see where does Sinex graph intersect intersect y equal to half graph. Isn't it that's the meaning of solving this equation. So Sinex graph is this white graph sinusoidal graph and half is y equal to half line. Which is horizontal parallel to the x axis. So these are the places which I'm showing with yellow. These are the places where the intersection of the yellow, sorry, the white graph of Sinex and the blue line of y equal to half happens. Okay, and there are many, many such points which will be existing. Right. So we will have infinitely many answers for this type of question. Because there are infinitely many types of answers or solutions to this question we normally divide the solutions into two categories. One is what we call as so solutions will be divided under two categories one what we call as the principal solution. Okay, principal solution. And other is what we call as the general solution. Okay, please note this down. Principal solutions I should write. Okay, what is principal solutions all solutions that belong to 0 to pi interval 0 to 2 pi interval, including 0 but excluding 2 pi. Now in many books I have seen in many books I have seen they have written it as 0 to 2 pi including 2 pi but please note that 2 pi will not be included it is 0 to 2 pi including 0 but not 2 pi. So all the solutions that belong to this interval those solutions will be called as the principal solution. Okay, for example, in our case, sine x equal to half. Can you tell me what are the angles possible between 0 to 2 pi excluding 2 pi for this sine x becomes half. You will say sir 30 degree. Correct. I hope everybody is aware. So 30 degree. So in our example, in our example, sine x equal to half. The possible value of principal solutions would be x equal to 5 by 6 which is 30 degree. I'm using radians to express the solutions. Then 5 by 6. Okay, these are the only two angles. So these two angles will be called as the principal solutions. Okay. So many times when you read the question also they will categorically ask you for principal solutions. Right. Now what is this general solution as the name itself says this will be a you can say a formula or a pattern which will try to give you all possible solutions for that trigonometric equation. Right. Because these functions are periodic, they have so many solutions. If I write a single standalone formula by which you can generate all possible solutions which will be infinitely many in number, then that formula would be called as the general solution. Right. So it's nothing but a pattern which represents or a formula or a rule by which you can find all your solutions to this equation. Now this is something which is slightly difficult and challenging. So we will learn this in this, you know, our chapter of trigonometric equations. Meanwhile, let us do this general solution finding for our given example sine x equal to half. Okay. Now all of you please contribute to this. Let it not be a monologue. Please start listing the angles for which sine x is equal to half. Let's let's do this exercise here. Okay. Now start giving me all possible values of x for which sine x becomes half. Of course, pi by six and five pi by six is already there. Can I write five pi by six like this? I hope that doesn't disturb somebody's understanding here. Anybody's understanding here. Okay. I'll come to your answer. Okay. What is the next value? Take and tell me sine x becomes half for what all values of the angle x pi by six. That is 30 degree. I'm sorry. Yeah. 30 degree and five pi by six. I've already mentioned. So whenever your angle comes to these positions, you'll start getting sine x equal to half. So see 30 degree. I've already covered. Okay. Then. Pi minus pi by six, which is 150. I already covered. Okay. Then I'll take complete round and come back to this position. Correct. Yes. Can I say this two pi plus five by six? I know seven pi seven pi by six will not give you the answer scope because at seven pi by six sign will become negative because you're in the third quarter. Okay. Yes. Yes. Keep telling me. Keep telling me. Okay. What is the next position? Next position is when you come here, that will be nothing but three pi minus pi by six. Am I right? Are you? Are you all with me? See, I started with pi by six, then five five by six, then full circle and coming back to the same position here. So you have to come keep coming to these two green lines position. Okay. So now this angle is two pi plus pi by six. And now that what I've shown you, that will become three pi minus pi by six. Okay. Tell me the next one. Tell you the next one. Then I'll think, then I will understand that you're, you know, your concept is clear. Okay. Next one is going to be four pi plus pi by six. Absolutely. Correct. Okay. Next one. Tell me. Next one. Tell me. Write down. Write down. Everybody write down the chat box. Next one. All right. Five pi minus pi by six. Very good. She won. Okay. Next one. Next one. Next one. Six five plus five by six. Absolutely right. Okay. Now see everybody. I'll just write a dot dot dot. Are you able to see a pattern in the solution? Are you able to see a pattern in the solution? In fact, you can go actually back also. For example, this pi by six, you can write it something like this. Zero pi plus pi by six. And you can go back also. For example, you can have, you can have something like. Yeah. Minus one pi minus five by six. That is also. Sorry. Minus one pi. Yeah. Minus one pi minus five by six. Okay. Yes. Now, if you look at the way these angles have been written. There is a pattern hidden here. What is the pattern hidden here? All of you, please pay attention. Whenever the coefficient of pi is odd number, there is a minus sign coming over here. Are you observing that? See odd number minus sign. Odd number minus sign. Odd number minus sign. And whenever there's an even number in front of pi, there is a plus sign. See even number plus sign. Even number plus sign. Even number plus sign. Right. Now, try to generalize this by a, by a formula. So can I say the formula that will suit this requirement is. N pi. Plus minus one to the power N. Pi by six. Right. Where N is a integer. Right. See, this is a very beautiful formula, which actually takes care of this particular pattern that you're observing over here. Whenever N is odd, automatically minus one to the power odd number will become minus sign. So a minus sign will automatically feature. And if N is even minus one to the power even number will become plus. Right. So automatically plus will come up. Yes or no. Yes or no. Okay. So this is the general solution for this equation that I was talking about. So as an example. Okay. For sine X equal to half the general solution. I'll be using GS for it going forward. The general solution will become. X equal to N pi. Plus minus one to the power and pi by six. Where N is a integer. So this becomes your general solution for this question. Is it fine? Any questions? Any concerns? Anybody has. Please do let me know. Okay. So now it is clear what is the difference between the two types of solutions. By the way, general solution is a, you can say superset and principal solutions are those values from the general solution which are under the interval zero to two pi excluding two pi. Is it clear? Any questions? Any concerns? Okay. I'll be doing many such activities with you, especially related to the general solution. Okay. And of course here and there we will also talk about principle solution. And then towards the end of this exercise that I'll be doing with you. I'll be giving you some formulas to take away from this entire exercise. Okay. So some learnings that we will do from this entire set of exercise. I'll give you some formula. First note this down. And if you have any questions, do let me know. Any, anywhere you want to copy, do let me know. You can, you know, ask me to shift the screen anywhere. I'll, you know, more than happy to help you. Okay. Clear? Should we go to the next slide then? Should we move on to the next slide? Let's target another example. Sinex is equal to, let me give you minus one by root two. Okay. Now my question to you is for this, let us find number one, the principle solution, the principle solution. And number two, the general solution. Okay. So one more exercise on similar to the type one question or the first type of question. Okay. So let's first write down the principle solution. Everybody, please give me the answer to the principle solutions. Which angle between zero to two pi will satisfy sine x equal to minus one by root two. Okay. Very good. Very good. So there is only one of the answers. There will be one more answer slope. Very good. Satyam. So sine x is negative. Are you sure? Okay. So sine x is negative in the third quadrant and in the fourth quadrant, right? Because in the third and the fourth, in the third only tan and cot can be positive in the fourth only cos and c can be positive, right? So think of an angle, a right slope. Absolutely, right? So think of an angle in the third quadrant and in the fourth quadrant for which, for which you will get sine x equal to one by root two with a negative sign. Negative sign means already you are in the third and the fourth. So just think about the magnitude. So now the magnitude here will be, please remember, it will be pi plus pi by four, which is nothing but, which is nothing but five pi by four. And the other one will be, I'm just making a small one here. The other one will be this blue one. Okay. And that would be nothing but two pi minus pi by four. That's nothing but seven pi by four. So absolutely, right? Those who said five pi by four solutions, let me write five pi by four and seven pi by four. Absolutely correct. Okay. Now principal solution is done. What about the general solution? So for this, I will start doing the same activity as what I did in the previous slide. Okay. So the very first angle, let me write down general solution as the name of the second question. Okay. So the first answer that comes to your mind will be five pi by four that you have already told me. By the way, yeah. Next answer that comes in your mind is, can I also write this like this? Five pi by four. Okay. I'll write it in the next step maybe. Yeah. Yes. Five pi by four next value. What does it come out? Anybody? In fact, minus pi by four also you can write it. Yes. Next one is two pi minus pi by four. Yes, I know. Okay. What are the next value? Keep moving. Keep moving in such a way that you land up on these two lines. Next value. Three pi plus pi by four. Very good. Next value. Four pi minus pi by four. Absolutely. You're in the right direction. Keep moving. Keep moving. That's why I wanted to write it initially itself, but now I can. Yeah. One pi minus pi by four. This is zero pi minus pi by four. Yeah. Next one. Next one. Next one. Yes. Five pi plus pi by four. Correct. Now here also I would request you. Enough, enough. We have done enough number of, we have enough data points to now know the pattern. So tell me what pattern do you see when you're writing it? So you see that. You see that you are basically following the same pattern. Okay. Yes or no. So let's say when N is even, you'll end up getting pi. Okay. When N is even, you let, let's say I take N as two. So let's say two pi. This will become minus one to the part two minus pi by four, which is as good as two pi plus minus pi by four, which is as good as two pi minus pi by four. That's what we had got over here. Okay. So what I wanted to show by the help of the second example is that you can actually generalize this and say, if you are asked to solve sine X equal to K. Okay. Where you could express K as sine of some known angle alpha. Okay. Then the general solution for this equation. Then the general solution for this equation is given by X equal to N pi plus minus one to the power N alpha, where N is a integer. So please make a note of this. Please, you know, keep this as one of your formulas in the formula list of your technometric. So knowing this formula, you can now solve many questions in a faster way related to this type of equation. So related to this type of equation, you can easily write down all the formulas in quick time. So let me just put a box around it. Yeah. So for example, if somebody asks you, let's say sine X equal to root three by two, find the general solution. So all you need to do is all you need to do is what is the general solution. So all you need to do is just write root three by two as sine pi by three. So just now I told you that this is the general solution. So you can write general solution as X equal to N pi plus minus one to the power N pi by three and belonging to integer over done. You don't have to sit and do all these, you know, listing of the angles to know the pattern. Okay. So initially it was good enough for us to, you know, do this exercise because we wanted to know the formula. Now that you have known the formula, you can directly make use of it to solve equations of this nature. Is it fine? Any questions anybody has? Okay. So one last example will take up and after that we will move on to the next set. So please solve sine X equal to negative half. Find the general solution for this equation. What do you think is the general solution for this equation? Give me an answer on the chat box. Sine X equal to negative half. What is the general solution for this equation? Very good Vishal. That's absolutely right. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Correct. Sine X becomes a minus half. One angle that comes to mind is minus pi by six. So I can write the general solution for this as X equal to N pi plus minus one to the power N minus pi by six. If you want it is your call. You can write it as minus one to the power N plus one pi by six also. Nothing will go wrong. Okay. So this is your general solution for this equation. Now here, one question that many people ask me, sir, could I take seven pi by six also because sign seven pi by six is also minus half. Yes, you can take any angle that you want. However, I have seen some books. They mentioned that this angle should be in the principal value branch of those functions which actually is not required. So you can take any angle that comes in your mind. For example, if somebody mentions somebody mentions this as science seven pi by six and writes the general solution to be x equal to n pi plus minus one to the power and seven pi by six. His answer is also going to be correct. This answer is also acceptable. This answer is also acceptable. Now many times you will be guided by the options sitting in the MCQ type question. So if MCQ type of question says this is the answer, then you have to mark this. Or if the MCQ options only contains this, then you have to mark this. So don't be very rigid that I have to get this in my options. No, they may write your options like this also. So even this is fine. See, ultimately what you want is when you list down your n is an integer. So when you list down all the answers from this yellow formula or this white formula, you are going to get the same infinitely long list of solutions. They will not differ from each other. Are you getting my point? Is it clear? So please understand here that there is not going to be any you can say marks deduction even if you write it like this or write it like this. So I saw Arya using this solution. So I absolutely write Arya that this is also correct. Of course, the n value that you are going to put here and the n value that you are going to put here, they might differ from each other. But ultimately the list that you are going to form will be same for both of them. Is it fine? Any questions? Any concerns? Any questions? Any concerns? See, I'll give you an idea. If you put n as zero here, let's say in this you put n as zero. What do you get the value of XS? What do you get the value of XS minus 5 by 6? Here if you put n as one, what do you get the value of XS? Check pi minus 7 pi by 6. What is that? Minus pi by 6. So ultimately the solution came out to be the same. But for the white one, you put n as zero. For the yellow one, you have to put n as one. Now, are you going to miss out any solution? No, because n is all integers possible and integers themselves are infinite sets. So they're not going to end ever. So eventually you are going to get all the answers that you should be getting. Whether you use the yellow formula or you use the white formula, there will be no difference. Got the point? So please watch out for the options. So which option to tick depends upon what options are there in the question. Let's do a similar exercise now for let's say this question. Cos X equal to let's say half. So let's find out number one, the principal solution, principal solutions and let's find out the general solution. So principal solution is very simple. Please give me a response on the chat box. For what angles between 0 to 2 pi do you think cos X is going to become a half? Fast, fast, fast, fast, fast, fast, fast. What are the angles? Pi by 3, absolutely. Minus pi by 3 is not in 0 to 2 pi. Absolutely right slope. So it will be 2, sorry, pi by 3. And 2 pi minus pi by 3. 2 pi minus pi by 3 will be 5 pi by 3. So these will be your principal solutions to this equation. Got it? All right. Now for the general solution, let us do some pattern recognition. So we'll start with pi by 3. Okay. 5 pi by 3, I will write it as 2 pi minus pi by 3. So in short, basically you have to come to these two positions. Either you can be in the first quadrant, pi by, oh, sorry. Yeah, yeah, pi by 3 or in this position, 2 pi minus pi by 3. So you already covered pi by 3. You have already covered 2 pi minus pi by 3. What are the next angle? 2 pi plus pi by 3, correct? Yes or no? Yes or no? Yes or no? Yes or no? What are the next one? What are the next one? Next one is back to this position, which is 4 pi minus pi by 3. Absolutely right. Absolutely right. Next one. Next one is this position. What is that? 6 pi, sorry, 4 pi plus pi by 3, correct? Okay. Next will be which position? Again coming back to this position, tell me which angle? Which angle? 6 pi minus pi by 3, correct? See, aria. Please understand from the beginning. Okay. This is pi by 3. Then this position is 2 pi minus pi by 3. Then this position is 2 pi plus pi by 3. Then again this position is 4 pi minus pi by 3. Then again this position is 4 pi plus pi by 3. Then this position is 6 pi minus pi by 3. Now tell me the next one. Next one will be this position. What will be that? 6 pi plus pi by 3. Okay. You have to come to this position aria, not to 8 pi position, this position. Okay. So now I think we have enough data points to figure out what is the pattern hidden? Can somebody tell me the pattern hidden? If you want, I can go back also. Let's say 0 pi plus pi by 3, 0 pi minus pi by 3 dot dot dot. Can somebody tell me the pattern hidden? What pattern do you see as the coefficient of pi? So 0, 2, 4, that's always even right. Absolutely aria. So can I say it is like 2 n pi? And now remember for both, for any value of let's say multiple of, even multiple of pi, you both have a minus and a plus. So you can say plus minus pi by 3. Is it fine? n being some integer n belonging to integer. Please note this down. Please note this down. Your general solution for this equation cos x equal to half will be 2 n pi plus minus pi by 3. Okay. So now I will save your and my time and let us generalize it straight away after one example only. So let me generalize it. So if anybody gives you cos x or cos of any variable equal to some k where that k could be written as cos of some known angle alpha. Then the general solution is given by, then the general solution is given by x equal to 2 n pi plus minus alpha n being integer. Please make a note of this. Please note this up. Is it fine? Any questions, any concerns? Anybody? Okay. So let's take one or two more questions based on the same. If I ask you write down the general solution of cos x equal to negative root 3 by 2. What is the general solution? What your answer will be? Write it down on the chat box. What is the general solution for this equation? Very good Vishal. Anybody else? Very good. Satyam. Harshita. Okay. See, first tell me any angle that comes to your mind for which cos becomes negative root 3 by 2. 150 degree. Correct. 150 degree is 5 pi by 6. Is it? So can I say the general solution for this will be 2 n pi plus minus 5 pi by 6 over n being some integer. Okay. By the way, some of you are giving me the general solution. Some of you are giving me the answer as minus 2 n pi minus pi by 3. Please note that minus pi by 3 cos becomes half. So cos of minus 60 degree is half. So it's not going to meet this requirement. Okay. So this becomes your answer to this question. Is it fine? Any questions? Any questions? Any questions, any concerns? Okay. Try one more. Try one more. Cos x is equal to, let's say, let's say, let's say, let's say 1 by or negative 1 by root 2. Find the general solution for this. What is the general solution for this equation? Very good. Very good. Satyam. Satyam. Slog. Anybody else? Harshita. Very good. Venkat, again, same mistake. Venkat, try to understand this. See, minus 1 by root 2 will be obtained when cos is either in the second quadrant or in the third quadrant. When you're writing minus pi by 4 here, you have taken it to the fourth quadrant. Fourth quadrant is where cos is positive. It cannot give you minus 1 by root 2. So either you say cos 3 pi by 4. Or you say cos 5 pi by 4. Both are fine. But you can't say minus pi by 4. So let's take this as our alpha value and write down the general solution. So general solution will be x equal to 2n pi plus minus 3 pi by 4 and belonging to integer. This would be an answer to this question. Is it fine? Any questions? Any doubts? Any concerns? Please do highlight. All right. Moving to the next example, let's say tan x equal to root 3. Okay. Let's write down the principal solution, general solution for this simple, basic, technometric equation. So let's start with principal solutions first. So think of the angles which come between 0 to 2 pi or which lie between 0 to 2 pi, including 0 but not 2 pi, for which your tan x becomes root 3. Write your values down on the chat box. Very good. Very good. Okay. So root 3 becomes a small mistake. Tan pi by 3 is root 3. So pi by 3, that's in the first quadrant and one more you will get in the third quadrant when you add a pi to this. So that will become 4 pi by 3. So these are your principal solutions. Tan pi by 6 will give you 1 by root 3. Got it? But our question was tan x equal to root 3. Okay. So 60 degrees and 240 degrees, which is your 2 pi by 3, sorry, 4 pi by 3, that becomes principal solutions. Now, the general solutions, for the general solutions, we need to observe the pattern. Okay. So let us start observing the pattern. So let me start with pi by 3 itself. And then 4 pi by 3 is like pi plus pi by 3. So the idea is, dear students, that in order to get the general solutions, we need to keep coming to these positions. One is here. Okay. Pi by 3, 60 degrees and to this position. Okay. So you have already accounted for pi by 3. You have already accounted for pi plus pi by 3. Now you have to come back to this position. So what will you write? Tell me. 2 pi plus pi by 3. Correct. Now you have to come back to this position. What will you write? 3 pi plus pi by 3. Excellent. You have started observing the pattern. Very good. Next, I have to come to this position now. 4 pi plus pi by 3. Absolutely. Okay. Now I think we have enough data points now to figure out what is the pattern involved. So what is the pattern involved? Please write down on the chat box. Right. N pi plus pi by 3 N being integer. Correct. So we will further generalize it. This is only for tan x equal to root 3. So now, since you all have understood this concept, let me save your and my time by jumping to the generalization of this particular equation. So in order to find the general solution of tan x equal to k kind of an equation where k could be written as tan alpha. Now many people ask me, sir, what if I am not able to figure out an alpha for which k is tan alpha? Then what do we do in that case? Right. So nothing to worry. I'll tell you the solution for that also. So the general solution for this will be the general solution for this will be x equal to N pi plus alpha where N is an integer. Okay. Note this down. Oh, sorry. Okay. Now starting from let's say the first formula that we did here. If you're not able to figure out an alpha, if you're not able to figure out an alpha, then you can also write x equal to N pi plus minus one to the power N sin inverse alpha in your answer. Okay. N being integer because many times they will give you unfamiliar numbers also. For example, if I give you, let's say something like one by six. Now, normally in our day to day, you know, dealing with tignometry, we don't know any angle for which sign gives you one by six. For that, probably we'll have to refer to the log book. Okay. A log table or maybe use a calculator, which is anyways not allowed. So in those cases, you can leave your answer like this N pi plus minus one to the power N sin inverse one by six. That's fine. Okay. But if you know the alpha, for example, if it is a known ratio, for example, half root three by two, one by root two or negative of these one zero, then of course you can put your actual value alpha. Is it fine? Okay. Similarly, here also I'll just include in case you are not able to find a known alpha, you can write it as plus minus cos inverse k sin inverse k only. I don't know. What did I write? Oh, sorry. Thanks. My mistake. I wrote alpha. Thank you. Thank you for correcting that. Okay. Yeah. And then cos inverse k. Fine. And I think in the last one, which I took just now, if your alpha is not, you know, you're not able to find out an alpha for which tan alpha is a K, then you can also leave your answer like tan inverse K. Okay. That is also fine. Is it okay? Any questions? Any concerns? Let's take a simple example. Further on this. Let's say if I say tan X equal to one by root three. Tell me the general solution. Tell me the general solution. Write down on the chat box quickly, quickly, quickly. Excellent. N pi plus now we are one by root three is a familiar ratio, right? For which the angle is 30 degree, right? Pi by six. So it's N pi plus pi by six. Okay. N being some integer. So this becomes your general solution for this question. Is it fine? Any questions? Any questions? Sorry for this trademark. I'll let it again. Not an issue. Yeah. Now, carrying forward with the same exercise. Now I would like you to tell me the principal and the general solution for this equation, sine square X equal to one by four. Write down the principal solutions and the general solutions for this equation. Let's focus on principal solution first. Yes. Tell me what are the values of X for this sine square X is one by four. Very good. Now everybody please pay attention. Okay. Okay. See everybody please pay attention. Sine square X equal to one fourth is equivalent to solving X equal to half as well as sine X equal to minus half. Okay. So you will get certain values of X coming from this equation, which is pi by six, five pi by six. And you'll also get certain values of X coming from this equation, which is going to be seven pi by six. Correct me if I'm wrong and 11 pi by six. Am I right? So four solutions are going to come for principal solutions are going to come for this equation. Okay. So these will be your principal solutions. Fine. Now, what about the general solution? What about the general solution? Let's try to figure that out. Okay. Let's see the pattern. Remember sine X equal to half or minus half will eventually cover all the four quadrants because in every quadrant, you will either have the positive value coming or the negative value coming. For example, in the first quadrant, in the first quadrant, you will have 30 degree coming in the second quadrant, you will have 150 degree, which is five pi by six. In the third quadrant, you will have seven pi by six. In the fourth quadrant, you will have 11 pi by six. Correct. So keep this in mind and start giving me what are the what are the possible angles that you are getting so that we can observe a pattern. So first angle is pi by six, which is this one. Correct. Let me make it in white. Yeah, this one. Okay. Next angle is five five by six that I will write it as pi minus five by six. Okay. Next angle is this angle, which is actually pi plus five by six. Next angle is this angle, which is going to be, which is going to be two pi minus five by six. Yes or no. Next angle is going to be this angle. What is that two pi plus five by six. Very good. Next angle is going to be this angle. What is that three pi minus five by six. Very good area. I'm almost starting. Yeah. Okay. Yeah. Next value will be this angle, which is three pi plus five by six. Right. See now I think you should have enough number of data points. Dot dot dot. What is the general solution or what is the pattern that you see here? The pattern that you would be seeing here is n pi right plus minus five by six and being integer and being an integer. Isn't it? Okay. Is it fine? So this becomes your general solution to this question. Okay. Now many people ask me this question and I think the same would be running in your mind as well. Why don't we solve this equation or why don't we write the general solution for this equation which happens to be x equal to n pi plus minus one to the power n pi by six? And why don't we write the general solution for this equation. Which is n pi plus minus one to the power n minus pi by six and why don't we do this? Why don't we take the union of those two solutions? Why don't we take the union of those two solution? x is equal to n pi plus minus one to the power n pi by six union with n pi plus minus one to the power n minus pi by six. Now, let me tell you this is correct. I mean, I'm not denying it. I'm not saying that this is a wrong result or something. But normally the question center, I mean, I have never seen a question where the question center has mentioned the answer like this. Okay. First of all, it looks very complicated. Okay. It is not wrong. Please don't get me wrong. I'm not trying to say this is a wrong result. No, by no means I want to say this is wrong. But when you have a simpler version of the same thing, why to use a complicated version, unless until it is there in your options. Okay, that I cannot guarantee that a question center cannot give this as an option. But if you can use or if you find this option sitting in your question, go for this option. You don't have to look for this option always. Okay. So this is also correct. But this way of writing the answer is very rare. I have never seen answers written like this. In fact, even if they have written it is very less in occasion. Okay. So this is an alternative way. Let me write it as all later. Okay. So this is an alternative way to get the same question done. But this is not a you can say preferred way of writing your answer. This is more preferred. This this fellow is more preferred. Are you getting my point? Any questions? Any concerns? So now can we generalize this as well? Yes. So let us generalize this as well. So if somebody gives you an equation like this, sine square x is equal to k, of course, a positive quantity. Okay. Let me write it like this k square, where you can write k square as sine square alpha alpha being some known angle, then you can write the general solution as then you can write down the general solution as x equal to n pi plus minus alpha. Or if alpha is not, you know, a well known angle, that means you are not able to figure out an alpha for such scenario, you can leave your answer like this also sine inverse. Okay. Of course, n being integer. Is it fine? Any questions? Any concerns? Put it on copied? Any questions? Okay. Let's take a simple example. Just to further get this concept here. Let's say if I ask you to solve sine square x is equal to half. Okay. Find the general solution for this equation. Very good. So this can be written as one by root two square. Right. One by root two square is sine pi by four square. Right. So you can write the general solution without much face of time n pi plus minus pi by four n being integer. Is this fine? Please don't get confused with this formula and the one which we wrote for tan x equal to k. Right. There there is no plus minus. There's only n pi plus alpha. So please don't get confused between this formula and this formula. Okay. There's only n pi plus alpha. Fine. The one which we took in the previous slide. Initially what happens? People tend to mix up those two. Okay. Is it okay? Any questions? Any questions? Any concerns? Fine. Now all of you please pay attention. Something very interesting that we are going to talk about. If somebody asks you to solve this kind of equation. Okay. Right. For example, let's say if somebody asks you, let me just write it down like this equal to let's say one by four. Okay. Which is nothing but cos pi by three the whole square. Correct. Now please understand here. Everybody please understand here. Saying this and saying this are the same things. Yes or no? You all agree? Okay. So writing this and writing this are same things. Right. Now just add a one to both the sides of the equation. So this becomes sine square x is equal to sine square pi by three. Correct. So this is as good as saying x equal to n pi plus minus pi by three. Which means that there is no change in the general solution even for this kind of an equation. Okay. So here please note everybody. Even if somebody asks you to solve, let me generalize this. Okay. Even if you somebody asked you to solve cos square x equal to k square where k could be written as cos alpha. In this case also the general solution is the same as what we saw in the previous slide. In this case also it will be n pi plus minus alpha. Okay. And being an integer or if let's say alpha is not a familiar angle, it will be n pi plus minus cos inverse k. Okay. Is it fine? Any questions? So a big sigh of relief that you have to now learn one formula less at least. Isn't it? Okay. So let's take an example based on this as well. I already took one example. I'll take one more. Cos square x is equal to three by four. Please write down the general solution for this. Please write down the general solution for this. Done. Okay. So three by three by four is nothing but root three by two the whole square. Correct. Which is nothing but cos of five by six. Correct. So cos pi by six whole square. So the general solution will be x equal to n pi plus minus five by six. Is it fine? Any questions? Any concerns? Is it fine? Okay. Great. Now let's talk about a similar set for tan also. Let's say if I ask you tan square x is three. Okay. Write down the general solution for this. Okay. Now all of you please pay attention. Saying this is as good as saying this. Correct me if I'm wrong. Correct. Or tan square pi by three. Let's write it like this. It is as good as saying. It is as good as saying reciprocal of this or you can say add a one to this. It is as good as saying this. Which is as good as saying this. If you reciprocate it, it is as good as saying this. And this is something which we have already figured out which was n pi plus minus pi by three n being integer. Okay. So what are the moral of the story here? The moral of the story here is actually it doesn't matter whether you take sine square x equal to sine square alpha cos square x equal to cos square alpha or tan square x equal to tan square alpha. For each one of them, the general solution will be the same. Okay. So even if we generalize this, even if we generalize this, so if tan square x equal to k square where k could be written as tan of alpha, okay. Then even for this, the general solution remains the same. So for the last three that we have done today, the formula is going to be same for all of them or same for the last three of them. Okay. Or let's say if your angle alpha is not well known, you can also leave your answer like tan inverse k. Okay. And being some integer. Is it fine? Any questions? Any concerns? Okay. So let us summarize. Or you want me to do, give you one more problem. Okay. Let's say one more problem before we summarize. Give me the general solution of tan square x equal to let's say, let's say, let's say one. Let's keep it simple. Yeah. Give me the general solution. Sometimes they will not say the word general solution. Sometimes they will just say the word solve. Solve this equation. So whenever you have a trigonometric equation and they have used the word solve, it is as good as asking you for the general solution. Yeah. Please write it down. Absolutely right, Satyam. So it's n pi plus minus pi by four n being integer. Okay. Because one is as good as tan pi by four the whole square. Is it fine? Any questions? Any concerns in the six formulas that we have done today? Yes. We have done six formulas in the last one hour of the class. So I'll summarize them also in the next slide. But before I move on, please let me know if you have any issues, any concerns, any questions. Okay. No doubt. Okay. So let me give you a summary. I mean, somebody doesn't mean the chapter is over. I'm just giving you a summary of whatever we have done till now. Okay. So if you have, if you have sin x equal to sin alpha kind of a scenario. Okay. By the way, just because I'm using x over here, that doesn't mean all the equations that you will come across, they will use x only. They could use theta also. Okay. They could use any variable they want. Okay. So for such case, for such case, let me just make a table kind of a scenario. This is my serial number. This is your equation. This is your general solution. Yeah. So for sin x equal to sin alpha, the general solution is x equal to n pi plus minus one to the power n alpha or sin inverse. I mean, if suppose this is not convertible to sin alpha, then you can write sin inverse key also and being integer. If you have been given cos x equal to cos alpha, the general solution is two n pi plus minus alpha. If your equation given to you is tan x equal to tan alpha, then your general solution is n pi plus alpha. And the fourth is basically, let's say if you have any of the three, sin square x equal to sin square alpha or cos square x equal to cos square alpha or tan square x is equal to tan square alpha. For all these three, the general solution that we follow is n pi plus minus alpha n being integer. Is it fine? Any questions? Okay. So this is a quick summary of whatever we have done in the last one hour, 15 minutes or one hour, 12 minutes of the class. Okay. Now let's start solving actual questions. Okay. Let's start solving actual questions. Let's start with this question. If sin alpha one and cos two alpha are in GP, find the general solution for alpha. Harshita, I can go back to that slide a little later on. First solve this question. Yes. Anybody with success? Okay, Vishal. Very good. See, what has been given that these three terms are in geometric progression. So when three numbers are in geometric progression, are in GP, we have already seen that in a geometric progression, B square will be equal to AC. Correct. So B square means one is equal to AC. AC means sin alpha cos two alpha. Yes or no? Yes or no? Okay. Now, can we do one thing? Can we write cos two alpha as one minus two sin square alpha? Right? From our half angles formulae? Okay, Vishal, you want to change your answer? That's fine. It's okay. All right. Now, if you please pay attention, this is as good as saying sin alpha minus two sin cube alpha. Now, this leads to a cubic equation in terms of sin alpha. Isn't it? This leads to this equation. Yes or no? Yes or no? Okay. So assume that your sin alpha is like your y. Okay, so let sin alpha is y. So it's like two y cube minus y plus one equal to zero. Can you? Now, this is a cubic equation. So we have to guess one of the roots in order to factorize it. So can you guess one of the roots? In my opinion, I believe minus one is going to work. Just check this out. Will minus one work? So two into minus one plus one plus one. Yes, it satisfies. So this works. Okay. So y equal to minus one satisfies. So this you can take y plus one to be a factor. So let me factorize it in a faster way. Since there is a two y cube sitting over here and there's only a white pulled out, there will be two y square coming and there is a one sitting over here and there is a one over here. So one will be there and in between there will be something like let's say ky. Let's find that k out. Let's compare this with this. So if you just take the coefficient of y square from both the expressions, there is no y square. So let me write it down comparing coefficient of y square. So in the top one, there is no y square. So you can write a zero and in the down one, you'll get a y square from this two into one and k into y. So k plus two is equal to this. So k value is going to be negative two. That means this is as good as saying y plus one times two y square minus two y plus one. So now please understand that this expression, this expression can never be zero. Why can this never be zero? Because you can complete a perfect square with that. Let me show you how. You can write it as twice y square minus y plus one. So you can write this as y minus half the whole square minus one by four plus one. Okay. I was going to give you two times y minus half the whole square plus half. So this term will always be greater than half. In fact, greater than equal to half. So it can never be zero. Okay. So this is the reason why this term can never be zero. And hence the only possibility is y is minus half. This is the only possibility, which means you're trying to say sine alpha is a minus one. Right? So if sine alpha is a minus one, minus one, you can write it as sine negative pi by two. So you can write the general solution as n pi n pi plus minus one to the power n minus pi by two. Okay. And being a integer, this becomes your solution to this question. Is it fine? So absolutely right, Vishal. I think Vishal got it absolutely right. Yes. Yes. Yes. You can write three pi by two as well. No issues. Instead of minus pi by two, you can write three pi by two as well. No issues at all. Is it okay? Any questions? Any concerns? Shall we go to the next question now? Okay. Sorry. I think Harshita wanted me to take back to the previous slide. Okay. Yeah. Yes. Harshita, if you want to copy something, you can do so. Fine. So we'll move on to the next question now. Okay. Let's take this one. Solve this equation, sine square theta minus cos theta equal to one by four and write the values of theta in this interval zero to two pi. Let me just add one more point to this question. Let's say we also want to find out the general solution. Let's say we also want to find out the general solution. Okay, Vishal. See, what you learned, they were the building blocks. Okay. So use those building blocks now to address slightly complicated problems like this. Now, when you see this question, what comes to your mind? What are the first thing that comes to your mind when you see this equation? Don't you feel like converting everything to cos so that you can make a quadratic equation out of it? Yes or no? Correct. And thankfully, I can do so because sine square is sitting over here. So I can write this as, okay, Harshita, Vishal, slow, very good. Let's discuss it out. Right. So can I write it like this? Yes or no? Yes or no? Okay. So let's simplify this even further. So four minus four cos square theta minus four cos theta equal to one. That means four cos square theta plus four cos theta minus three equal to zero. Any questions? So this is a plain and simple quadratic in terms of cos theta. Okay. So let cos theta be y. So you can just treat it like a quadratic four y square plus four y minus three equal to zero. Is it fine? Can we factorize this? I think so. We can easily factorize this. Okay. Take a two y common y plus three. And I think take a minus one common y plus three equal to zero. So this is easily factorizable as two y minus one times y plus three equal to zero. Okay. Now put your cos x, sorry, cos theta back. Okay. Oh, I'm so sorry, even at this position. Right. So this gives us two possibilities. Either two cos theta minus one is zero. That means cos of theta is half. Can we write a general solution for this? Can we write a general solution for this? Yes. So general solution will be two and pi. Now half is as good as half is as good as cos pi by three. Right. So it will be plus minus pi by three as per the formula that we have seen a little while ago. And the other possibility is actually not a possibility because cos theta cannot be, or let me write it like this, cos theta is equal to minus three by two, which is not possible, which is not possible. Cos of theta cannot be beyond one or cannot be lesser than minus one. So this is ruled out. So this is the only possibility. This is the only solution possible. Okay. By the way, they were asking for zero to two pi. Okay. So zero to two pi. Let us solve this. So if your theta has to lie between zero to two pi, the possible set of solution will be pi by three, of course. And I think five pi by three. Okay. So these two will be your zero to two pi solutions. Is it fine? Any questions? Any concerns? Is it fine? Can you move on to the next question? Okay. Let's take another one. By the way, all of you are expected to be good in your basic trigonometric identities. So here is a question which says solve cos theta plus cos theta plus cos five theta plus cos seven theta equal to zero. By the way, the meaning of the word solve is finding the general solution. Okay. So solve just means finding the general solution. Oh, good. So this will act like a revision for you. Okay. Vishal, anybody else? Okay, Satyam. Okay. So let's try to solve this question. Let's try to club these two expressions together and these two expressions together. So I hope all of you remember the formula of cos A plus cos B. Cos A plus cos B is two cos A plus B by two into cos A minus B by two. I hope all of you have retained the reduction, sorry, a transformation formula in your mind through practice. So the first two terms will give you two cos theta plus three theta four theta by two, which is two theta and theta minus three theta, which is minus two theta divided by two, which is minus theta, but cos does not care about negativity of the angle because cos is an even function. So whether you say cos minus theta or cos theta, it doesn't care about the negativity. Same with this also, it'll become two cos six theta into cos theta and this is given to you as a zero. Take two cos theta common, you will have cos two theta plus cos six theta. So this will become two cos theta. This is going to become two cos four theta into cos. If I'm not mistaken, this will be two theta again. Yes or no? In short, it simplifies to something like this. So now everybody, please pay attention. If the product of three quantities is zero, either this could be zero or this could be zero or this could be zero. Correct? Now the general solution for this will be theta equal to two n pi. Please note that zero means cos pi by two, right? So two n pi plus minus pi by two. Now all of you please pay attention. Something very important I would like to discuss over here. When you say two n plus minus pi by two, if you see here, if you put start putting dummy values for n, let's say if I start with n as zero, that will give you theta value as pi by two and minus pi by two, correct? When you put n value as a one, that will give you theta value as now see two pi minus pi by two will be three pi by two and two pi plus pi by two will be five pi by two, correct? Similarly, if you put n as two, you will get theta value as five pi by two and seven pi by two. Sorry, n value as one, n value as two will give you four, yeah, sorry, seven pi by two and nine pi by two. I'm so sorry, seven pi by two and nine pi by two. Okay? And so on. By the way, if you see it is catering to all odd multiples of pi by two, so this solution can be bettered by writing it like this, two n plus one pi by two or you can write two n minus one pi by two also both are fine, okay? So generally speaking, you will not see this written in your answer many a times, right? Instead of that, they will write either this or this, okay? So when cos theta is zero, you can just say it's an odd multiple of pi by two, that's it. See, I'm not saying that this is wrong to write or this is the only way to write it. No, I'm not trying to say that. Both are acceptable, both are fine, but why to write it like this when you can write it just two n plus one pi by two and odd multiple of pi by two? Simple as that. Similarly, for here, I can write two theta as two n plus one pi by two, which means theta could be two n plus one pi by four n being integer. And this last one, this last one, you can write it as four theta as two n plus one pi by two or theta is equal to two n plus one pi by eight n being some integer. Is it fine? Any questions? So three set of solutions are possible here. What are the possible solutions? Theta could be two n plus one pi by two or two n plus one pi by four or two n plus one pi by eight. Okay, so these are the possible set of solutions. Okay, any questions, any concerns? Please do let me know. So I think Satyam, you only got, I think one of these three, right? So all the three are possible. All the three are possible. Okay, any questions? Any questions? Please note this down. And if you have any questions, please feel free to highlight it out. Let's try to slightly complicate these questions. Can I go to the next slide? If you are all are done, can I go to the next slide? Okay, let's take this question. Who do the power one plus mod cos x plus mod cos x the whole square plus mod cos x cube and so on till infinity is four. Find the general solution for this equation. Okay, Satyam. See, first of all, on the numerator, if you see the power is a geometric progression, correct? Right? As you can see, it's an infinite geometric progression. Correct? So this is a infinite geometric progression. Am I right? Infinite GP, right? Now remember two to the power something is giving you four. That means this four is actually two to the power of two. That means this power is a two. Now for an infinite GP, let's recall for an infinite GP, what is the sum? What is the sum of an infinite GP? Of course, you know, the number of sorry, the common ratio of this term should be less than one a by one minus r. Absolutely. So this will become one by one minus mod cos x. Right? In other words, one minus mod cos x is half. That means mod of cos x is half. So dear all, please remember, this is as we're saying cos square x is equal to half square, which is actually cos pi by three squared. So remember the general solution for this was n pi plus minus pi by three. Please don't forget those basic formulas. So this becomes your solution to this question and pi plus minus pi by three. Is it fine? Any questions? So Harshita was absolutely right. Very good, Harshita. Is it fine? Satyam also got it right. Okay. Let's discuss now some important facts, some important type of questions. Let me first start with this type of question, which is normally asked in your competitive exams. a cos x plus b sin x equal to c, even if you can call it as a sin x plus b cos x also equal to c. Does it matter? So how do we solve this kind of a question a cos x plus b sin x equal to c? Okay. Now here, please understand a few things. Before I start solving this type of questions, there are a few things I would like to write down for you all. Let's understand a few things over here. I'll come back to this topic name. So let us understand important points we remembered for solving any equation. In fact, these are some general guidelines we should all follow while solving equations. Okay. After this, I'll come back to that type. Okay. So the first thing is, the first thing is avoid squaring both sides of any equation. Now, please note the word. I'm not saying don't square. I'm saying avoid. Avoid means as far as possible, don't square both sides of the equation. But however, there are some equations which you cannot solve without squaring. So you have to square. Right. So why do we ask people not to square unnecessarily is because squaring introduces extraneous roots into the system. So extraneous roots will creep into the system. Okay. So this leads to extraneous roots. Extraneous root means extra roots will come into the system. Okay. So for example, if somebody has this equation x equal to 1. Okay. And he tries to square both the sides. Then this will lead to this will lead to this scenario. Okay. And unnecessarily he'll get two answers, one and a minus one, whereas the solution was only one x value was only supposed to be one. But because you squared it, you ended up getting one more answer, which is actually an extraneous solution. Okay. Now I'm not trying to say by any means that don't square and solve it. If you have a necessity that without squaring, I will not be able to solve it. Then go for it. But mind you, after the process is over, that means after you have got the solution, you must check which of the solutions that I have got does not satisfy my original equation. Those are extra roots. They have to be removed from your answer. Right. Are you getting my point? So square it and solve it where you have the liberty or where you have the, you can say, you know, the easiness to actually figure out or actually remove the extraneous solutions from your answer list. Okay. So here is a word of advice. Don't try to square where there is a general solution being asked because if a general solution is being asked, even the extraneous roots will be very, very large in number. So removing all those solutions would become a tedious task, a complicated or complex task for you all. Okay. I'll tell you what do I mean by that. If there are a few in number, for example, if you have been asked to find the solution in a limited value of x, that's a zero to two pi or zero to four pi, then you can square and solve and whatever extra roots come into the system, you can check with the original equation and remove those extraneous solutions. But wherever general solution is asked, don't square it because if you square and solve it, even the extraneous solution will be infinitely many in number. Removing them will become a Herculean task in that case. Okay. I'll come back with an example on that. Second thing that you should keep in mind is do not cancel, do not cancel an unknown factor. This I've been telling this since very beginning. Okay. What does this lead to? This leads to loss of genuine solution. Okay. The most simplest example that I keep giving people is let's say x square is equal to x. Now, what are you doing? You're cancelling one of the x's and you write x equal to one. Please note that this equation could have zero also as one of its solution. So x could be zero also and x could be one also, but you lost this guy because you just scored off xx from both the positions, both the sides. So do not cancel any factor which is unknown to you. Unknown to you means, I don't know what is x, and still I'm cancelling it out. Okay. So please cancel out factors which is whose value is known to you. For example, if there was a 2x and there was an let's say 4x square, I would have cancelled 2 and 2 because I know 2 is not zero. Okay. Or let's say somebody tells you that okay x is not equal to zero. Then you can cancel it out. But don't cancel any unknown factor because that factor could have been a zero and then it could have been a root of that solution which otherwise you will lose on cancelling. Okay. So this is something which I have told you since the Briskow stays. So don't cancel any unknown factor. It will lead to loss of genuine solution. Okay. Third thing, do not accept any solution that makes any function or any sub function involved. Let me write sub function involved in the equation involved in the equation, undefined. Okay. Undefined. Okay. So please do not accept any solution that makes any sub function or functions involved in the equation as undefined. In other words, no solution beyond the domain is acceptable. Beyond the domain is acceptable. Okay. Please understand this. See, let me give you an example. Let us say you were given some equation. Okay. And that equation contained let's say tan x. Fine. And you solve that equation somehow and you got one of the answers as pi by two. Now, will you accept pi by two as your answer? Given that there was a tan x in the equation, will you accept pi by two? You'll say no. Because at pi by two tan x is not defined. So many times it happens, there are so many trigonometric identities that we convert one into the other and we solve it and we realize that, okay, I got an answer. But if that answer is making the original equation, any function undefined, any function, even if one of the function is undefined, that solution will be rejected. You will not be honoring that solution. You will be just rejecting it. Are you getting this point? So please keep these things in your mind whenever you are solving any equation in this world need not be only trigonometric equation, any equation. For example, if you're dealing with something like under root, okay, the equation contains under root of something. So let's say you've got an answer for which that under root thing is becoming, I mean, the expression within the root is becoming negative. That answer has to be rejected, isn't it? Because you can't operate, you can't apply an even root on a negative value that will give you imaginary answers. Are you getting what I'm trying to say? So whenever you're solving an equation in whatever way you are solving it, go for it. I'm not restricting the way you like, you would like to solve an equation. But make sure the answers that are coming, okay, they make the original equation, you know, they satisfy the original equation. Okay, they should not make any term undefined. What's the point? Okay. Now with this, we are going to move on to the type which I was going to talk about because it involves, one of these important points that I have given you. So I should have actually given you this in the beginning of the first question. In fact, before the first question. So let's say this is the equation that we normally get to solve in many competitive exams, A cos x plus B sin x equal to C. Okay. Now, let me illustrate this with an example. Let's say somebody gives me this equation to solve. Now you tell me, how would you solve this equation? So let's solve this equation under two situations. Number one, solve when your x is between 0 to 2 pi. And second, solve for the general solution or get the general solution. Okay. So I've given you two scenarios to solve. Okay, one is your answer should be between 0 to 2 pi. That's a very, you know, you can say limited value of x which I need. So your answer should be only between 0 to 2 pi. I don't want beyond it. And the second part of the question is I want the general solution for it. Okay. Now all of you, please pay attention. Please pay attention. If you have been given a very restricted value of x from where you have to find the solution, you can actually go for squaring as well. That means I could go for squaring both the sides. So for the first one, I could go for the squaring. Now you would be thinking, sir told avoid squaring as far as possible. Yes, avoid it. But in this case, squaring is going to help me out to find the solution faster. Okay. But mind you, my dear students, this will come with extraneous solutions. Why this will come with extraneous solutions? Because this incorporates not only the solution for our given equation, but this also will, you know, additionally give you solutions for this equations also. Isn't it? So this equation we don't want. We only want this equation solution. But when you square it, your solutions for both the equations are going to come under the same. That means extraneous, extraneous, extra solutions are going to come up. Getting my point? Never mind. Since it is a limited range, I'm going to easily, you know, remove them off. How I will show you that as well. So when you square it, you get cos square x plus sin square x plus 2 cos x sin x equal to 1. Right? Now cos square plus sin square is anyways a 1. And by the way, 2 cos x sin x is as good as a sin 2x. Isn't it? Yes or no? That means sin of 2x is a 0. Correct? So sin 0 is also 0. So you can write down the general solution as 2x is n pi plus minus 1 to the power n into 0, which is as good as n pi. Which means x is n pi by 2, n could be any integer. So our solution for 0 to 2 pi could be 0. So when I put n as 0, I will get a 0. It could be pi by 2. It could be 2 pi by 2. It could be 3 pi by 2. Correct? Or it could be 4 pi by 2, which is 2 pi. Correct? Now see everybody, please pay attention. Does 0 satisfy our equation, original equation, which is cos x plus sin x equal to 1? Is 0 satisfying the equation? Tell me yes or no? Yes. It is accepted. Correct? Is pi by 2 going to satisfy our original equation? Yes. Is pi going to satisfy the original equation? Think carefully and say no. This guy is an extraneous solution. How did it come up into the system? Because of that squaring activity. So because I squared, it led to pi creeping into my solution set. Are you getting my point? But it is easy to identify because this range itself is a small range. 0 to 2 pi only. I can figure it out. 3 pi by 2, will it satisfy it? No, because cos 3 pi by 2 is 0, but sin 3 pi by 2 is negative 1. So it is basically satisfying the second equation, whose solution I don't want. 2 pi will it satisfy? Yes. So there are only three solutions possible. So your x value will be either 0, pi by 2 or 2 pi only. Is it clear any questions? This was the first part, when your solution is only between some very restricted interval specified by the question center. But when it comes to general solution, please don't square it because here you could handpick them and remove them. Let's say your mom is cooking rice and that rice has got some small stones into it. It's very easy to remove those stones if there are few. But let's say half the rice is stone only. You'll say I'll rather not use this rice at all. Because you will waste so much of time picking those infinitely many number of stones present in it. So if your rice has a limited number of stones, you can consider handpicking them and throwing them off. But if your rice contains infinitely many stones, you'll rather not use that rice at all. So the same thing will happen in general solution also. In general solution, those number of stones, which is an unwanted stuff, they will be infinitely many in number. So why to use this method? So the second solution that I'll be showing you, I will not square it. So what I'm going to do, see very interesting. Rather, I'll go to the right side of this thing. So for the general solution, what I'm going to do, I'm going to use a concept which I had discussed with you in the trigonometric function chapter, which is called the harmonic form. So what I'm going to do, I'm going to create something like this, all of you please pay attention. So this term that you see here, it's actually nothing but cos x minus pi by 4, am I right? So basically I'm converting it to a single cos function. You can convert it to sine function also, it is your call. But many times it depends upon the options also. Is everybody convinced with the fact that the original equation that we had can be written like this? Yes or no? Which means cos x minus pi by 4 could be written as 1 by root 2. 1 by root 2 is cos pi by 4. So can I write a general solution for this like this 2n pi plus minus pi by 4? Which means my x can be written as 2n pi plus minus pi by 4 plus pi by 4. Okay, n being an integer. But again, this is not a good way to write the answer. So you could take two cases out of it, one with a plus sign and other with a minus sign. Okay. So the first one will give you 2n pi plus pi by 2. And the second one will give you 2n pi. Okay. Now all of you please pay attention. So the general solution will become your x could be either of the nature 2n pi or of the nature 2n pi plus pi by 2. Now check. If you put some values, let's say I want to know how many solutions are possible between 0 to 2 pi. So let us use this general solution to get the answer to the first part of the question. So 0 to 2 pi, I can choose n as 0. So if I choose n as 0, I will end up getting x as 0 and pi by 2. Okay. And if I put n as a 1, I will end up getting 2 pi. And now remember this will exceed 2 pi, right? Because when I put n as 1, this will become pi pi by 2. So this is the only three solutions possible. 0 pi by 2, 2 pi. Is it matching with our answer that we had got? 0 pi by 2, 2 pi c. Okay. So I got the same answer, but this time I did not square it. Why did not I, why I didn't square it? Because if I square it, it will lead to lot of extraneous solutions. And those lot of extraneous solutions, I will not be able to remove. Okay. So this is the way to solve it without squaring as well. Is this fine? So please note this down. Then I'll give you some generic, you know, gyan about this a little later on. First, note this down. You want me to drag the screen up and down? Let me know. Now, I know many of you would be, you know, having this query in your mind how to, you know, convert this whole thing, you know, easily like this. That also I will revise with you how to do it easily. Okay. So just copy this down, then I will tell you how to do it for the generic cases. Now, I would prefer converting it to cos actually rather than sin. See, I have an option to convert this expression in terms of sin also and cos also both options I have, but I would prefer converting it to cos because cos general formula is less complicated. Isn't it 2n pi plus minus alpha, right? Sin general formula is slightly complicated n pi plus minus 1 to the power n alpha. So I prefer converting it to cos rather than sin. But again, this is not a compulsion. You can use, you can convert it to sin and cos any of the two that you want. And that also is decided by the options many a times. Okay. Can you scroll down? Yes, why not? This place, is this fine? Done? Okay. Now, let us write down the general procedure to solve it. Let's say if somebody asks you to solve this equation. Okay. So what I do normally is I write a, now I'll tell you how to convert it to a single cos function also. That trick also I'll tell you. Write this a as r cos theta. Okay. And write this b as r sin theta. Okay. And get your r and theta by simply squaring them and adding them. So if you square them and add them, you will get r square. By the way, cos square plus sin square is going to become a 1. So let me just write it. And you can get your r as under root of a square plus b square. Now, both the possibilities are there minus plus. Take plus. Okay. That is good enough for you to solve the question. No need to take another one, minus. And you can get your theta by taking the ratio. So divide second by the first. So if you divide second by the first, you get tan theta as b by. So your theta value also can be obtained very easily. Okay. So for example, in our previous case, we had cos x plus sin x. So a is 1, b is 1. So in this case, r will be 2. For this case, r will be 2. And theta will be tan inverse 1, which is pi by 4. Okay. Now, once you make these substitutions here, this is what you're going to see. Take r common, you get cos x cos theta plus sin x sin theta. Okay. Everybody, by this time, you know, this is your compound angle identity, which is cos x minus theta. So just put this r as under root a square plus b square. Okay. Put your theta as tan inverse b by a. Okay. That is something which you already did in that previous question. So that is why the previous question became root 2 cos x minus pi by 4. So your cos x minus tan inverse b by a becomes c by under root of a square plus b square. Right. Now, this is a place where you can now write this guy as cos of some angle alpha. Right. And you can write the general solution to be this. So your x will become 2n pi plus minus alpha plus tan inverse b by a. Okay. N being some integer. So this is your final answer to this question. Now, you don't have to remember this result. Okay. Many people, they tend to remember this also, but I don't want you to put unnecessary burden on your memory because you have many other things to remember. Okay. I think inorganic chemistry itself is going to take vast amount of your memory space. So as a maths formula, I don't want you to remember this. This you can solve in the runtime also. Okay. Is this fine? Any questions here? Please note this down. By the way, again, I'm telling you, you could have converted this to sin also and done it. So instead of a as r cos theta, you could have taken as r sin theta also and b as r cos theta also and solve the same problem. Okay. So cos and sin that depends upon options. What options to take? What options are given so that you can decide which process or which method to take while solving it? Is it clear? Any questions here? Okay. Now a few pointers to be noted. All of you please focus on this step. Okay. This step tells you that cos of some angle is c by under root of a square plus b square. So from this step, I can say for solution to exist, for solution to exist, this term c by under root of a square plus b square should actually be between minus one to one. Why? Because it is cos of something, right? Cos of any angle, no matter whatever is the angle, is restricted between minus one to one. In other words, c should be between negative under root a square plus b square. Okay. In other words, your mod c should be lesser than equal to under root a square plus b square. So please make a note of this. Very important. For solution to exist, mod of c should be under root of, should be lesser than equal to under root of a square plus b square. That means while solving such an equation, if you realize that oh, my c, mod c is not less than under root a square plus b square, then straight away say there is no solution to that equation. Don't waste your time solving it. Are you getting what I'm trying to say? See, for example, if somebody says sin x plus cos x equal to two, let's solve this equation. Okay. So here, see a value is one or whatever, b value or a value, whatever you want to call it, doesn't make a difference. So let's say this is your b value, this is your a value and this is your c value. Do you realize that mod of two is not less than equal to under root one square plus one square? Correct? Because mod two, which is two is actually more than under root two. So for such equation, there is no solution. Don't waste time solving it also. Don't waste even a single second solving it. We are not going to talk about any complex solutions. That is why I know in the beginning itself, I told you there should not be any solution which is making your function undefined or non-real in nature. Okay. No complex solutions. Even if the question setter doesn't mention, please understand it. Our solutions are always real in nature. We are not going to talk about complex solutions, non-real solutions. Okay. Unless end in the question setters mentions it explicitly. Of course, in quadratic equation and all, they would not write it. But while solving such equations like trigonometric equations, algebraic equations, we always write the real solutions. Okay. If you want, I can put a real word over here for real solution to exist. That would make it more precise. Right, Satya? For real solution to exist, mod of c should be lesser than equal to a squared. Now, this is very important because a lot of questions have been framed on this itself. Okay. We will take some questions and then we'll take a break. I know there's a break time already. So just one question we'll take and then we'll take a break. Okay. Let's take this one. Find the general solution. Solve means find the general solution for this equation. So first of all, ask yourself, is this equation solvable? Will I get a real solution for this? Ask yourself. That means is mod c less than equal to under root of a squared plus b squared? Then only go forward. So if you compare this with a cos x plus b sin x equal to c, you realize that a square plus b square under root. So a square. So a is your, sorry, a is your root three. So a square will be three. Okay. Actually doesn't matter the order. b square is one under root. Is c lesser than equal to this? The answer is yes. The answer to this is yes. That means the solution exists. So real solution does exist. Okay. Exist. Now let's go ahead and solve this. So what do I normally do? I follow the same procedure as what I discussed in the previous slide. I take root three as r cos theta and one that is your coefficient of sin x as r sin theta. Squaring and adding gives you r square as four. That means r is two. Take the ratio. Tan theta is one by root three. So theta value is pi by six. So when you put it back, when you put these two back in your equation, you get r cos theta cos x r sin theta sin x equal to root two, which is nothing but r cos x minus theta equal to root two where r is two theta is pi by six. So this is a simple equation to deal with. So cos pi x minus pi by six is one by root two. One by root two can be written as cos pi by four. That means x minus pi by six is two and pi plus minus pi by four. In short, x is two and pi plus minus pi by four plus pi by six. But normally when we write it, we break it up and write it like this. We write it as two and pi plus pi by four plus pi by six and then take a negative two and pi minus pi by four plus pi by six. So this boils down to two and pi plus five pi by 12 and this boils down to two and pi minus pi by 12. So here is your final solution. So your general solution will be two and pi plus five pi by 12 or two and pi minus pi by 12 and being integer. Is it fine? Any questions? Any concerns? Make sense? Okay. So here we'll take a small break. Okay. On the other side of the break, I'll take one more question based on the same type and then I will show you some other types as well. So let's take a break. Right now as per my laptop, it's 621 pm. We'll meet exactly after the 15 minutes break, which is 636 pm. Okay. Normally during the break, even I take a break. So the camera and the mic will be turned off. The recording will be stopped. So you can go enjoy your break and come exactly at the same time back. Okay. See you on the other side of the break. So before we left for the break, we had taken the type A cos x plus B sin x equal to C type of a problem. I would take just one more question on the same and then we will go to the next type. So let's take this question. So let's take this question. Yeah. So this question says if k cos x minus 3 sin x is equal to k plus 1. Okay. If k cos x minus 3 sin x is equal to k plus 1 has real solutions, has the real solutions, then which of the following option is correct? Option number A, k has to be greater than equal to 4. Option B, k has to be less than equal to 4. Option C, k has to be between 0 to 4. Okay. And option D, k could be any real number. Okay. So which option is the most appropriate option for this question? I'll put the poll on for you all. Okay. I'll put the poll on for you all. So please mark the most appropriate option for this question. I hope my writing is legible. A cos x, k cos x minus 3 sin x is equal to k plus 1 has a real solution. Then k satisfies which interval? k cos x minus 3 sin x equal to k plus 1 has real solutions. Then which is the most, which is the most appropriate option which satisfies this? Okay, Vishal. If anybody is not able to see the poll, he or she can also put your response on the chat box. Okay, Shlok. Very good. Three people have responded on the poll so far. Okay, Satyam. I can give one more minute. So those who are trying hard, please try to wrap this up in one minute. It's actually less than 30 seconds question. But still I'm giving you around two minutes for this. No worries, Harshita. You can correct yourself on the chat. Okay. You can mark, you can tell me the option which you think is correct. Okay. All right. So let's stop now. 5, 4, 3, 2, 1, stop. Okay. I could see a mixed response. A, B, C have got two options each. D has got one option. Okay, let's see which option is going to be the correct one. So as I already discussed with you for a solution to exist, okay. I'll rewrite the situation here. A cos x plus B sin x equal to C. For a solution to exist, mod C should be less than equal to under root of A square plus B square. So this is the condition for solution to exist. Okay. So for a real solution to exist, this is the condition that we need to satisfy. Okay. So here our A is K. B is minus three. So A square plus B square under root. This should be less than mod C. Mod C is mod K plus one. Now you can square both the sides. If you square both the sides, this is what you're going to see, which is as good as saying K square plus 2K plus one is less than equal to K square plus nine. K square, K square gone. So 2K is less than equal to eight. So K is less than equal to four, which means option number B is correct. Okay. So option B is the right option. Is it fine? Any questions? Any concerns? Any questions? Any concerns? Okay. Next type that we are going to discuss now before I move on. Any questions, anybody? Oh, no problem. Okay. Next type that we are going to take is the type where it involves a homogeneous function. So it's a homogeneous... So the equation involves a homogeneous function in sin x and cos x. Okay. Let me give you an example for it. So equations involving homogeneous function in sin x and cos x. Let's take an example. Let's take this question. Okay. If you see this question, this term is actually a homogeneous expression in sin x and cos x. Now, what are the meaning of homogeneous expression or homogeneous function in sin x and cos x? If you look at each one of these terms, each term will be having the same power effectively on sine or cos. For example, here, there is a power of two. Here, one sine and one cos is multiplied. So again, the effective power is two. Here also, cos square x, effective power is two. So you can see that there's a uniformity in the power of sine or cos. Now, many people say, sir, right side the power is zero. So how is it homogeneous? See, even if it is four, you can actually write it as four sine square plus cos square. So the power is, what are the meaning of the word homogeneous? Homogeneous means there's a uniformity. I mean, I'm sure in chemistry, you would have studied homogeneous mixtures. So the quantity of the solute and the solvent is same in any part of the mixture. In the same way, in any part of the equation, you will see that the effective power on each of these terms is actually uniform or homogeneous. That is why it is called homogeneous function. That is why these expressions are called homogeneous functions in sine and cos. So how do we solve such kind of a question? So in order to solve this question, we normally convert this to a quadratic in terms of tan. So what do we do here? Let us first write this expression once again. So here what I'm going to do, all of you please pay attention. First, I'm going to bring these two terms to the left side. Let's see what happens because of that. It makes it sine square x and this becomes I think 12 cos square x. So what do we do now? We divide by cos square x so that it becomes a quadratic in tan. So divide by cos square x. So when you do that, the first term will become tan square x, second term will become 7 tan x, third term will be as good as a 12. So this happens to become a quadratic in tan. So is it factorizable? You can call tan x to be y for the time being if you want. So it's y square minus 7y plus 12 equal to 0. It is factorizable. I'm sure everybody knows how to factorize this y minus 3, y minus 4. So that means y could be 3 or y could be 4, which means tan x could be 3 or tan x could be 4. So if tan x is 3, what is the general solution? If tan x is 3, what is the general solution? x will be equal to n pi. Tell me, complete it, n pi plus tan inverse 3. Now there is no such angle that is known to us for which tan is 3. At least it is not one of the 0, 45, 60, 30, 90, etc. So you can leave your answer like that. n being an integer. And if you take tan x as 4, then the general solution would be x equal to n pi plus tan inverse 4. Got the point? So these two will become your answer. Is it fine? Any questions? Any concerns? Please do highlight. Is it fine? All right. So I think you will be able to manage any such questions. So if let's say there was a power of 3333 everywhere, let's say there was a homogeneous function of degree 3 involved everywhere, you divide by cos cube, get a cubic in tan. And in most of the cases, that will be factorizable. Don't worry. You will not get a cubic equation which is beyond your factorization. I can say capabilities. You will always get equations which are factorizable. So there's no point wasting time doing more and more questions. So let us quickly move on to next type. Next type is the equation which involves extreme values. So equations involving extreme values. Now what do I mean by this topic name? Equations involving extreme values. Let me give an example. Let me give an example. Let's say I want to solve this question. Now read this question very, very carefully. It involves three variables x, y and z. So you would be wondering, sir, one equation, three variables, how are we supposed to solve it? Because there's only one equation and there are three variables, x, y, z. So how do I solve all these three variables? Or how do I find the general solutions to these three variables? Now here, everybody take a minute and analyze this equation deeply. You will get the answer to this question. So the question that is coming in your mind, you will yourself find answer to that. Has anybody found the breakthrough? Has anybody found the breakthrough? No? No breakthrough yet? Okay. See, we all know that sin x is between minus 1 to 1. So can I say sin square x will be between 0 to 1? Same goes with cos y also. Cos y also lies between minus 1 to 1. So cos square y will lie between 0 to 1. And everybody please pay attention. Seek lies between now. Seek is an angle which lies between minus infinity to minus 1, union 1 to infinity. So seek square z will always be greater than equal to 1. Please understand this. So seek in square of any angle, no matter whatever is the angle, that will always be a value which is greater than equal to 1. Okay. Now all of you please pay attention. This guy is always less than equal to 1. This guy is also less than equal to 1. So overall this whole thing is less than equal to 2. Yes or no? This fellow will be greater than 1. So overall this fellow is greater than equal to 2. So your left-hand side is less than equal to 2 while your right-hand side is greater than equal to 2. And you're equating it. So when do you think the equality will hold true between left-hand side and right hand side? See again I'm repeating the scenario. Your left-hand side will always be less than equal to 2. Your right-hand side will always be greater than equal to 2 and you're equating it. So when do you think they will agree with each other? Right when both of them are equal to 2. Absolutely right. So your LHS and right-hand side both should be equal to 2 each. Then only the equality will be true. So for each to be 2 that means sin square x should also be 1. Cos square y should also be 1 and cx square z should also be 1. Then only it will be true. Correct. Now sin square x is equal to 1 is as good as sin square pi by 2. So what is the general solution? N pi plus minus pi by 2. Is it correct? This is as good as saying cos square y is equal to cos square 0. That means this is equal to n pi plus minus 0 which is as good as n pi only. Yes or no? This also is as good as saying cos square z is equal to 1. So z is also equal to n pi. Is it fine? Any questions? Any concerns? So see since it was an extreme value scenario you could find three variables just from one equation itself. Any questions? Any concerns? Exactly. And let me tell you Harshita and others that such type of scenario is seen very commonly in competitive exams especially KVPY is known to put such scenarios. We'll take one more of the same kind. Don't worry. We'll take one more. Okay. Everybody has copied. Any questions? Any concerns? Okay. Let's take the next question based on the same type. Extreme value scenario type. The question of that type. Okay. Nevertheless, I'll just put a question for my side. Let's take this question. Roo 2 to the power of sin x plus cos x to the power of 1 plus sin 2x is equal to 2. Find the general solution for this equation. Find the general solution for this equation. Sin x plus cos x whole raise to the power of 1 plus sin 2x is equal to 2. Find the general solution for this equation. Okay, Vishal. That was quick. Yes. Anybody? Okay, Satya. Very good. Okay, Harshita. Okay. Let's discuss it. See, all of us know that sin x plus cos x is a value which is going to lie between... Now, let's call this as a C. Okay. Let's say I call this as a C. Now, we all know that the C value is something which is between under root of a square plus b square, right? Which means mod C is less than root 2. In short, your C has to be between negative root 2 to root 2. Okay. In short, this guy that is sin x plus cos x will be some value which is going to lie between root 2 to negative root 2. Right? Agreed? Everybody? Now, everybody, please pay attention. This fellow is a value which will lie between minus 1 to 1. Right? Now, I have a proposition over here. The proposition is in order to become a 2, it can only happen when your sin x plus cos x either becomes a plus root 2 or a minus root 2 and the power of it becomes a 2, which means sin 2x becomes a 1. So, only when these two conditions are simultaneously met, when these two conditions are simultaneously met, simultaneously met, then only this equation is going to get satisfied, yes or no, yes or no. Agreed? Right? So, now, this could be clubbed as a single equation saying sin x plus cos x, the whole square is equal to root 2 square because it will cover both the scenarios plus minus root 2, which means sin square x plus cos square x plus 2 sin x cos x is equal to 2, which means sin 1 plus sin 2x is equal to 2, which means sin 2x equal to 1. So, both of them now actually speak the same thing. Right? So, you don't have to worry about two different equations. It is actually both our same equations. So, all you need to do is solve this equation. Okay? And this is as we were saying sin pi by 2. So, your 2x will be nothing but n pi plus minus sorry n pi plus minus 1 to the power n pi by 2. So, your x will be n pi by 2 plus minus 1 to the power n pi by 4. So, this is going to be your general solution to this question. Is it fine? Let me check how many of you got this right? Aiyo, people have missed out n pi by 2, I believe. Am I right? Satyam, Vishal, did you miss out that 2 down here? Yeah, both Satyam, I think Vishal missed it out. Is it fine? Any questions? Any questions? Any concerns? Please highlight. So, you have to analyze the situations. Many a times we just start solving it blindly. So, ask yourself, is this equation going to be satisfied for some extreme conditions? Right? Because many a times such questions are difficult to solve on its face value. On its face value, it looks very difficult question. Right? So, just ask yourself, is it like the question setter has put an extreme situation in order to make this equation satisfied? And that will come with practice. As you practice more and more questions, you will become more and more refined in identifying such cases. Getting my point. Okay. Next are those equations. So, another type. So, those equations which involve all the type of technometric functions, sine, cos, tan, everything is involved. Okay. So, how to solve such kind of a scenario? Again, I will like to solve this by an example. Okay, let's take an example. I think I could take this question. Now, first of all, I would like to give you a fair chance to solve this question. I would not like to jump on the theory straight away because that will kill the learning part from your side. So, first of all, I would request everybody, take a minute, take a two minutes. Okay. No problem. And try to see how are you going to solve this question. Everybody, please note that this into this plus two is also there. Okay. Don't ignore this plus two. Don't start comparing them to zero, zero, ignoring that two. Okay. So, there is a two there. See how you can solve this question. So, ideally, you can see this is a function of all these sine, cos, tan, etc. So, how can we go ahead and solve such a problem? Normally, what happens if there is a sine and a cos sitting there? We try to square both sides or we try to write a harmonic form for it like how I taught you using r cos theta and r sin theta. Many times we try to convert one in terms of the other if one is squared up by using our trigonometric identities. But when everything is involved like sine, cos, tan, as you can see, all the three functions are involved over here in some or the other form, then how do you manage such scenarios? Any idea? Anybody is actually making an attempt to solve it or have you got stuck somewhere? Okay. Anyways, so let me bail you out of that situation. See, when your function involves mixed trigonometric function like this, there is actually a method by which you can write this entire equation as a single trigonometric function and that root is by converting them into half angles of tan. Yes. So, the root is converting them to half angles of tan. See how sin x can be written in terms of half angles of tan like this. I hope all of you recall your trigonometric identities. Cos x can also be written in half angles of tan, something like this. Okay. And of course, tan x itself can also be written as half angles of tan. Right. So what is the benefit of converting it to half angles of tan is that you get a uniformity in, you get a uniform expression in terms of tan x by 2. So you can take tan x by 2 as a y and solve it like just any other polynomial equation. Right. So that is the benefit of this. So this basically method proves to be very, very useful when it comes to such mixed type of trigonometric equations. Are you getting my point? So what I'm going to do, I'm going to use this and I'm going to also use tan x by 2 as a, let's say, you know, you can call it as a y or something. Okay. So let's try to solve it. So I'm going to use, I'm going to use these three formulas in this left hand side. So cos x, cos x will be 1 minus y square by 1 plus y square minus sin x. Sin x is 2 y by 1 plus y square. See here, look at the formulas over here, which I've written in yellow. I'm just calling tan x by 2 as a y and I'm using it over here. This becomes 2 into 2 y by 1 minus y square and this is 1 by this, 1 by this will be 1 plus y square by 1 minus y square. Okay. Plus 2 equal to 0. Let's try to further simplify this. Let's try to further simplify this. So this is going to become, if I'm not mistaken, 1 minus y square minus 2 y. This is going to be 4 y by 1 plus y square and down in the denominator I'll have 1 plus y square, 1 minus y square, which is 1 minus y to the power 4. Any questions? Any questions? Let's take the LCM and let's take the LCM and send the denominator to the other side. So let's simplify this. So as for me, I believe that y to the power 4 will be the maximum power that will be occurring because I can see a minus y to the power 4 here and minus 2 y to the power 4 from there. So that will become minus 3 y to the power 4. Please correct me if I'm missing out anything. So y to the power minus y to the power 4 minus 2 y to the power 4 will give you minus 3 y to the power 4. Okay. Will there be any y cube term? Check. Yes. So I think y to the power 3 term will be coming from these two correct and these two. So if I'm not mistaken, that will give you minus 6 y cube. Am I right? Okay. What about y square term? Will I be getting y square term? Yes. I'll be getting y square term from here, from here, from here. So that will be giving you minus 8 y square because minus y square and plus y square that will get cancelled with each other and you'll be only left with minus 8 y square. Sorry. I'm just erasing this part. I'm rewriting it. Don't worry. Yeah. What will be the y terms? Y terms will come when this multiplies with this. This multiplies with this. Okay. So that will give you if I'm not mistaken plus 2 y and constant terms will be, I think just coming from here to this 3. So 3 will be your constant term. If you want, you can multiply throughout with a negative sign also and just write it like this. Now, this is a y quadratic in y. How do I solve it? See, since you have now, this is again a trick which I'm giving you. Okay. Many times it works. Since you put tan x by 2 as a y, think of those values of y which can be obtained from tan of certain known angles like y could be 0 in fact, but in this case, 0 will not work. It could be 1 by root 3. It could be minus 1 by root 3. It could be root 3. It could be minus root 3. So these are some generic values that we know for tan. Okay. Now, this is just a, you can say starting point for you. I'm not claiming that always one of them will work. No, don't get me wrong here. But as a person who is, let's say, you know, smart, he will try to put values which are obtained from tan of some angles like 1 minus 1, 1 by root 3, minus 1 by root 3, root 3, minus 1 by root 3. But I don't claim that these values will always work. Okay. See, you cannot solve a bi-quadratic equation of this nature like you're, you know, using your quadratic equation formula. Right. You need to do some guessworks. Okay. So my first guesswork will be does one satisfy it? Let's check. One will not satisfy it. The reason being, if one is to satisfy the coefficient should add up to give you 0, which is not happening. I hope all of you know the divisibility. If one is a root, or one is a root of this equation, all the coefficient should add up to give you 0. Right. That is not happening. Even minus 1 is not a root because if minus 1 is a root, the even ones should add up to give you the odd ones. Sorry. The odd ones, which is not happening. Now let's check 1 by root 3. Does it work? Let's check. So if I put 1 by root 3, this will become 1 by root 3 to the power 4, which is 1 by 3 square. This will become 6 by 3 root 3. This will become 8 by 3. This will become minus 2 by root 3 minus 3. Let's check whether it is working. So I think this is 2 and this whole thing will get answered. This is 1 by 3 plus 8 by 3 minus 3. Oh, wonderful. This works. Okay. So that means y could be 1 by root 3. Right. Now check for minus 1 by root 3. So minus 1 by root 3, let me just drag it up. Oh, there's no place down. Yeah. So let me put minus 1 by root 3. Minus 1 by root 3 will give you 3 into 1 by 3 minus 6 by 3 root 3 plus 8 by 3 plus 2 by root 3 minus 3. Again, these two get cancelled. Okay. And again, this will become 1 by 3 plus 8 by 3 minus 3. Yeah. That also satisfies. So y could be minus 1 by root 3 as well. Okay. So out of four roots, two roots I've already found out. Right. So let us find out the remaining two roots. Okay. But for finding the remaining two roots, I will use a simple strategy. Since y minus y could be plus minus root 3, it means y square is 1 by 3. That means 3 y square minus 1 is a factor. Isn't it? So you can, you can factorize that term, which was this whole thing. 3 y to the power 4 plus 6 y cube. Let me write it down first. So 3 y to the power 4 plus 6 y cube. I have to drag and copy plus 8 y square minus 2 y plus 8 y square minus 2 y minus 3. Yeah. So let us factorize this. So one of the factors is already this. So let us guess what is the other factor. By the way, other factor is going to be a quadratic, right? Because overall the degree has to be 4. Okay. So 2 degrees already there. So this has to be quadratic. Now, see how, how, you know, in a very short way, I'm going to get these coefficients. See, there is a 3 y to the power 4 and there is a 3 y to the power, sorry, 3 y to the power 2 and there is 3 y to the power 4. That means here y square should come. Correct? Yes or no? There's a minus 3 here and there's a minus 1 over here. So plus 3 should come. Correct? And in middle, there will be some term which is let's say KY. All I need to do is find my k first of all. So in order to find k, you can compare the coefficient of, let's say, y square on both the sides or y on both the sides. Anything is fine. In fact, y cube also can be compared. So let's compare the coefficient of, compare coefficient of y cube on both sides. See, by the way, you can do a long division method also. I'm not denying it. I'm not stopping anybody from doing long division method. Okay. You long divide this term by 3 y square minus 1. Get your answer. That is also fine. So if you compare the coefficient of y cube on the left side, you get 6. On the right side, you'll get 3 k. Okay. I don't think so. There's any other term that will contribute to a y cube, 3 k. So k is 2. That means this term is nothing but 2 y. Now, what are you trying to solve? We're trying to solve this equal to 0. So if you're trying to solve this equal to 0, it will result into only this being 0 because this cannot be 0. This fellow cannot be 0. Why? Because it is y plus 1, the whole square plus 2. This cannot be 0. So the only solution possible is what we obtained. So you will be thinking, I use that we wasted time, final realizing that there's no further route. No, it's not a waste of time. Okay. Please realize that it's better to be safe and get that four marks completely in your j main rather than taking this half hazard chances. Okay. So be safe. Don't take chances. So your tan x by 2 is going to be 1 by root 3. Let's write down the general solution. So tan x by 2 is tan pi by 6. So x by 2 is n pi plus pi by 6. So x is 2 n pi plus pi by 3. Okay. N being a integer. Next, tan x by 2 is equal to minus 1 by root 3. So without much waste of time, I will let it as 2 n pi minus pi by 3. Correct me if I'm wrong. Okay. So these two are your solution to this question. Is it fine? Any questions? Any concerns? So keep this also in your mind. See, I'm just exposing you to different types of questions, dear students. Okay. You should be prepared. You should have all the weapons in your armory. You never know which weapon will help you. Are you getting my word? So as a, you know, as a coach, I would like to show you all the facets that you can get under this particular, you know, chapter. So you should be ready that, okay, this comes, maybe this option can also be exercised on this way. This method can also be used. Any moment. Okay. Any questions, any concerns? Please do let us know. Now, let me expose you to more such situations. So I would like everybody to work out this question. Now, before you jump into solving the question, okay, just see whether you can convert it to a polynomial equation with some smart substitution somewhere. See, this is what, see, J is not about conventional methods. Oh, I know this type of question that is, that is, you know, every question of that type I will solve by using, you know, a conventional method. No. First, you need to analyze, see whether there are some loopholes or there are some, you can say weak points in the question which you can leverage. Okay. For example, in this question, I see a sign term raised to a very heavy power 10. Okay. I see a costum also raised to a very heavy power and I see a cost 2x. Okay, raised to a power of four. Is there any way by which we can write them in terms of a single expression? And of course, not involving a very high polynomial power. Can I do something like that? Any, any process that comes in your mind? Anybody, any process that is coming in your mind? Cost 2x is there, sine x and cos x is also there. Can you do something with it? Okay. See, many of you would be thinking of writing sine square x as a t. Correct? Because if you write sine square x as a t or a y, whatever, then cost to the power of, of course, sine to the power 10 could be written as y to the power 5. Okay. And even cost to the power 10x, you can write it as 1 minus, now I love you please pay attention, 1 minus sine square x to the power 5, which is as good as 1 minus y to the power 5. Yes or no? And not only that, cost 2x is actually 1 minus 2 sine square x. So that also you can write it as 1 minus 2y. Isn't it? So this to the power 4 can be written as this to the power of 4. So of course, you can write this whole thing as y to the power 5, 1 minus y to the power 5 equal to 29 by 16, 1 minus 2y to the power 4. Okay. Yes or no? This is one way to do it, but I will tell you a more smarter way. Okay. Of course, you will simplify, you will get a fifth degree equation. By the way, it will not be a fifth degree equation, it will be a fourth degree equation because y to the power 5 and when you expand it here, you will get minus y to the power 5. That will get cancelled off. Okay. But I will tell you a simpler way to solve this question. This is one way. Second way is, I know that sine square x is 1 minus cost 2x by 2. And similarly, cost square x is 1 plus cost 2x by 2. Okay. Yes or no? So what I'm going to do, sine to the power 10x, I'm going to write 1 minus cost 2x by 2 to the power of 5. And cost to the power 10x, I will write it as 1 plus cost 2x by 2 to the power of 5. Okay. So as a result, what I can do is, I can take cost 2x as a y. So your left hand side will become 1 minus y by 2 to the power of 5. 1 plus y by 2 to the power of 5. That will be left hand side. Okay. Right hand side will be 29 by 16 y to the power of 4. Yes or no? Now, this would be a simpler one to deal with. Let me tell you why. Let's multiply with 2 to the power 5 throughout. So multiply with 2 to the power 5 throughout. This will become 58 y to the power 4 because 2 to the power 5 is 32. Now, this term that you see, you will realize that most of the terms will start getting cancelled here. Let me write this separately for you. Let me first write 1 minus y to the power 5. Now, please remember binomial theorem. A little bit binomial theorem we already know from our bridge course days. But if you expand it by binomial theorem, this is what you are going to see. Similarly, this term is going to be 1 minus 5y plus 10y square minus 10y cube plus 5y to the power 4 minus 5 to the power 5. So when you add them, when you add them, the left hand side term is going to give you two times. See, this two will get cancelled. This two will get cancelled. This two will get cancelled. So that will give you two times 1 plus 10y square 5y to the power 4. So this has an add-on advantage that you will end up getting a y quadratic but that y quadratic will always have even power on y. So that is basically beneficial to us. How it is beneficial to us? Let me just explain this first of all. So this will become 24y to the power 4 minus 10y square minus 1 equal to 0. So if you take y square as a k, then this just becomes a quadratic, isn't it? Yes or no? And I think this will be factorizable. You can factorize it as minus 12k plus 2k minus 1 equal to 0. Let's take 12k common, 2k minus 1. So 12k plus 1 times 2k minus 1 equal to 0. Yes or no? Is it fine? Any questions, any concerns that you have till this stage? All good? Check it out and let me know if I missed out anything or any concerns that you have. Okay, so this is as good as saying 12y square plus 1 times 2y square minus 1 is 0. Remember, this cannot be 0. That means only possibility is this is 0, which means y is equal to plus minus 1 by root 2, which means cos 2x is plus minus 1 by root 2. Now I'm sure you can easily write down the solution for this. Let me write it down here. Yeah, this is as good as saying cos square 2x is equal to half. Okay, half is cos pi by 4 whole square. Okay, so that means 2x is equal to n pi plus minus pi by 4. So x is equal to n pi plus minus pi by 8 and belonging to integer. So this is going to be your answer to this question. Is it fine? Any questions, any concerns? Please do highlight. Okay, so we'll stop here. By the way, there are a few more types of questions that are left to be covered. And of course, we have not yet started with trigonometric in equations. So in our next class that we meet, we are going to, in our next class that we are going to take up, we are going to complete that part. I think one hour of class is more required to finish off this. And then I will be starting with properties of triangles. As I told you, next class properties of triangles will be taken up. Okay, so please note down, this is your final answer n pi by 2 plus minus pi by 8. All right. Thank you class. Bye-bye. Take care. Stay safe. Good night. See you in the next session. Bye-bye. Take care.