 So under trigonometric equation, we have to start solving certain miscellaneous type of trigonometric equation. So I call them as miscellaneous equations. The first equation that we are going to talk today is an equation which involves almost all trigonometric functions, the basic standard trigonometric functions, sine, cos, tan, c, cos, etc. So equations involving such kind of functions, how do we solve these kind of questions? See up till now the type of equations that we were solving in those equations, there was only one trigonometric function involved. Or if at all there was let's say two trigonometric functions involved, then one could be written in terms of the other. Like if you remember there was a question involving sine square, theta and cos theta, so we could write sine square in terms of cos. And if at all we had let's say both sine and cos which were not interconvertible into each other, then we use the harmonic form to convert them as a single trigonometric function. If you recall there was a form that we did a sine x plus b cos x equal to c. So that was a form where we converted it to a harmonic form. Now here in this case, you realize that you have multiple trigonometric functions involved. How to solve those kind of functions? Let's begin with a question. Let's begin with a question. Let's say this question. As you can see in this question, first read this question, you have cos x minus sine x times 2 tan x plus 1 by cos x plus 2 equal to 0. So all the type of functions are involved here, sine, cos, tan. So now how do I solve such kind of question? What is the standard operating procedure for solving such kind of a problem? So for such kind of a problem, we basically use the half angle formula in terms of tan. Because all the trigonometric functions, whether it is sine, whether it is cos, whether it is tan or even the reciprocals, they can all be written in terms of half angles of tan. For example, sine x could be written as 2 tan x by 2 by 1 plus tan square x by 2. Cos x could be written as 1 minus tan square x by 2 by 1 plus tan square x by 2. And of course, tan x itself can be written as 2 tan x by 2 by 1 minus tan square x by 2. So what is the advantage here? The advantage here is that instead of dealing with diversified kind of trigonometric equation, you are converting it to a single type of trigonometric equation, which is tan x by 2. So using a single trigonometric ratio to express this entire left hand side. So when you express the entire left hand side in terms of a single trigonometric ratio, which in this case is tan, then you can easily take that as a t. For example, I can take tan x by 2 as a t. So this entire expression will get converted to a polynomial kind of an equation. And polynomial kind of an equation is easy for us to solve. So let us try to apply this procedure onto this question and let's see how effective it becomes in solving the problem. So what I'm going to do is I'm going to write, I'm going to call tan x by 2 as a t. Okay, I'm going to call tan x by 2 as a t. So this will become 2t by 1 plus t square. This will become 1 minus t square by 1 plus t square. This will become 2t by 1 minus t square, correct? So if I substitute it over here, let's see what do I get? Let me put a barricade over here. So cos x will become 1 minus t square by 1 plus t square. Sin x will become 2t by 1 plus t square. 2 tan x will become 2 into 2t by 1 minus t square. And reciprocal of cos will become 1 plus t square by 1 minus t square plus 2 equal to 0, right? Plus 2 equal to 0. Now let's try to simplify it further. So numerator here will become t square plus 4t plus 1. And denominator will be 1 plus t square, 1 minus t square, which is nothing but 1 minus t to the power 4, okay? Plus 2 equal to 0, right? Let's multiply throughout with 1 minus t to the power 4. By the way, let me write it like this. Yeah, let's multiply throughout with 1 minus t to the power 4. So this boils down to this equation, okay? Now let's simplify. So here you have a t square. Here you have a minus, sorry, t to the power 4. Here you have minus 2t to the power 4. So that will become minus 3t to the power 4, correct? So see, minus t to the power 4 from here and minus 2t to the power 4. So that will make it as minus 3t to the power 4. Next, t cube coefficient will come from here minus 2t cube and minus 4t cube. So that will become minus 6t cube. t square will be minus t square plus t square and minus 8t square. So that will become minus 8t square, okay? And t will be 4t minus 2t, which is going to be plus 2t. And constant terms would be 1 plus 2, which is 3. Is it fine? Now you must be thinking, are you, sir? You ended up getting a bi-quadratic equation. How are we going to solve a bi-quadratic equation? Because we have only learned how to solve quadratic equations. Okay, so don't worry. Whenever you have any polynomial equation, I mean whose degree is more than 2, we have to guess one of the roots, okay? But there are methods to solve cubic and bi-quadratic equations also directly by using some, you know, you can say higher level concepts called the Cardin's method and the Ferrari's method and the Distartus method and all. But of course, right now we are not going to go into those concepts. So now we have to guess one of the roots here. At least one of the roots we have to start with, okay? Now, if you have taken your tan x by 2, now this is some, you know, gyan which I'm giving you. So tan x by 2 is taken as a t, right? Now think of some values which tan can give you. So normally tan is known to give, you know, values like 1 minus 1, however it can give any real values. But the ones which are commonly seen, okay? So if you talk about 0 degree, 30 degree, 45 degree, 60 degree, etc. The values which are commonly seen is something like 1 minus 1, 1 by root 3, minus 1 by root 3, root 3, minus root 3, etc., okay? So first, first try out such substitutions. Now, this is not with surety that one of the substitution will definitely work, okay? This is just a, you can say smart guess which I'm doing. It may fail, okay? I'm not trying to say that these substitutions will definitely work fine, okay? So since tan has been substituted as t, we will try with those values which are commonly seen in the ratio of tan, right? 1 minus 1, root 3, negative root 3, 1 by root 3, negative 1 by root 3. Now it may not work also, but that is the only option that we can try with, okay? Now, does 1 work? I don't think so because the sum of the coefficient should become 0 if 1 has to work, okay? So 1 is not going to work, correct? So this is not going to work. Is minus 1 going to work? No, for minus 1 to work, the sum of the even powers coefficient should add up to give you the sum of the odd powers coefficient. That means minus 11 plus 3 which is minus 8 should be equal to minus 4 which we know is not, correct? So even this will not work. Try 1 by root 3, does it work? Let's check 1 by root 3. So 1 by root 3, this will give me minus 3, 1 by 3 square, minus 6 by 3 root 3, minus 8 by 3 plus 2 by root 3 plus 3. Does it become a 0? Let's check. This is minus 1 by 3, minus 2 by root 3, minus 8 by 3 plus 2 by root 3 plus 3. I think this gets cancelled, and minus 9 by 3 which is 3. Oh yes, it gets cancelled, gives you 0. So this is working fine. This is working fine. Yes or no? Okay. So this is working fine. Try with negative 1 by root 3. Negative 1 by root 3 will also give you minus 1 by 3 plus 6 by 3 root 3 which is this. Minus 8 by 3, minus 2 by root 3 plus 3. Does it work? Oh yes, minus 9 by 3 plus 3. That's the RTFC was 0. So this also works. Okay. Now two of the roots out of four we have already guessed. Okay. ROOT 3 will it work? Check it out. ROOT 3 will it work? Minus 3, root 3 to the power 4 will be 3 square, minus 6 into 3 root 3, minus 8 into 3 plus 2 root 3 plus 3. 18 root 3, 24, no. This is not going to work. Okay. And I don't think so minus root 3 will also work. So these two are not going to work. Okay. But nevertheless, I think it's a great help that I have got two of the roots. Okay. What if it doesn't work? Then it's a difficult question to solve because I know solving a bi-quadratic equation is not something that is very easy. Okay. But such a situation will be very rare. So as to say, because the questions it knows that the question that he has, he or she has to provide must be solvable. Right. So of course he can give you, you know, weird and values to solve, but that's not how the questions are designed. Just to tell you when questions are designed, the question setters themselves sit and solve it. Okay. Just to talk about your school exams. Okay. Your school teacher first herself will solve the question paper. Do you know that? No. She has to prepare a marking scheme or the answer for that question paper. Okay. So the question paper is designed by the school management. So let's say I said a question, but I'll sit and solve it. So these, all the testing agencies, whether you talk about NTA or need people, they will solve the question paper. And when they realize that, no, this is something with the child will not be able to compete within two to three minutes. They will automatically remove the question. Okay. Trust me, this happens. They don't give orbit numbers. They don't give orbit figures. Okay. They will be challenged in the court of law. They will say, okay, solve this question in four minutes. You are expecting people to solve it in four minutes. You solve it in four minutes. If they themselves don't solve it, they cannot ask that question. Okay. All right. Now, anyways, so we have got two of the roots. So we have to find the remaining four roots. Now, all of you please pay attention. The equation which was given to you has T equal to one by root three and T equal to minus one by root three as two of the roots. So can I say root three T minus one and root three T plus one, they will be two of the factors. So can I say these are our factors of the left-hand side expression. So what was that like left-hand side expression? Okay. This expression, by the way, you can drop the factor of minus also. You can just write three to the power four plus six T cube plus eight T square minus two T minus three. So this expression that you have, this expression that you have, which I'm showing with the curly braces, that equation has got two of its factors as root three T plus one and root three T minus one. Okay. So in short, in short, three to the power four plus, if I'm not mistaken, it was six T cube, eight T square minus two T minus two T minus three. This can be factorizable as, see what is the product of these two? Three T square minus one. Okay. And some quadratic will come over here. Okay. So this question mark will be some quadratic, right? Because total number of roots are four, two you have already found out. So two more are supposed to be found out, which can be hidden in a quadratic expression. Okay. So how do you find this quadratic expression? I have a simple technique here. And that technique is also known to most of you. There is a three T to the power four sitting here and you already have three T square. So you can have a T square over here. So that the product becomes three T to the power four. Okay. You already have a minus three here. You have a minus three here and you already have minus one. So there will be a plus three here. Correct. Now the only thing which is not going to be, you know, easily seen is your coefficient of T. So you can assume it to be some KT. Right. So what I'm doing, I'm trying to factorize this without actually using the long division method. Okay. So I'm trying to factorize this without using a long division method. So what I'm going to do here is, what I'm going to do here is I'm just going to compare coefficient of, let's say TQ or T square or T any of the thing you can basically choose. Let's compare the coefficient of TQ from both the sides. Okay. TQ from both the sides. From both sides. So left side coefficient of TQ is six. Our right side TQ coefficient will come from three K. That's it. I don't think so. Any other term will contribute TQ. So K becomes a two. That means this expression is nothing but three T square minus one into T square plus two T plus three. Okay. Now what are we trying to solve here? You're trying to solve a bi quadratic polynomial equation. So instead of putting this as zero, you can afford to put this expression as a zero. Correct. Now remember this guy cannot be zero. Why? Why this guy cannot be zero? Why? Tell me. Write down on the chat box. Why this fellow cannot be zero? Tell me. So T square plus two T plus three can be T plus one, the whole square plus one, right? So this cannot be zero. So only possibility is three T square is equal to one. So T is equal to plus minus root three. That's the only possibility. Yes or no. Correct. Now. So your tan square X by two is one by root three, the whole square. Isn't it because T is nothing but tan X by two. So basically I've squared both of them, which is as good as saying tan five by six, the whole square. So what are the general solution for such kind of equation X by two is n pi plus minus pi by six. Yes or no. Which means X is two and pi plus minus pi by three. That's your answer. What is it? Do you explain yourself while solving a question in the exam? You say, okay, see to understand this is the way to solve it. Do you say all these things? Do you do all the steps which I'm showing you? No, right? So I'm explaining you how to do the problem. That is why it is taking so much time. If I have to do the problem, see what other steps I'll be writing. I'll be directly writing these steps. Okay. So these are the only steps which I'll be writing. Okay. So this question can be done in four minutes. Maximum five minutes, not more than that. Fine. Does everybody remember all the basic formula that we had done in the last class? If not, please refer to your notes. I hope everybody remembers how to solve tan square variable equal to tan square some angle. Right? I hope you remember this. Does everybody remember this or not? Yes, no, maybe. Okay. Great. Is it fine? Any questions here? So again, repeat the process. The process is if you realize that there is a trigonometric equation given to you where the trigonometric equation involves different types of trigonometric ratios, which are not readily intraconvertible. So that means you cannot make a single, you know, trigonometric ratio equation like what we did in one of the questions before we converted it to a quadratic in costs. Right? Or we cannot write it as a single harmonic form. Right? So in such cases, you can go with such methods. So this is one of the weapons in your armory. You can use it if there is a need like that. Fine. Is it okay? All right. So now we will move on to different types of, you know, another type I'm going to talk about. So second type that we are going to talk about our functions, which involve something of this nature. That means there are equations which involve sin x plus cos x or sin x minus cos x along with sin x cos x. Okay. So here also one of the substitutions is going to help you out. But before that, I'll give you the question itself where such a thing is required. So let me just take a small question. Yeah. Let's say this question. Now, first of all, you apply your mind. If you were given this question from the knowledge that you already had, how will you solve this question? Please note there is no sin square and cos square here. Okay. So if you are thinking that I'll divide by cos square and I'll start getting a quadratic in tan. That is not going to happen. Okay. This is a single sin sin x term, single cos x terms. This is the product of sin x cos x. Okay. So now here, what is the approach? All of you, please listen to me. Okay. So what do we do here? We take sin x plus cos x term to be T. So our effort is to convert it to a polynomial kind of an equation. So I will take sin x plus cos x to be a T. Now, what is the benefit of this? See, if you square both the sides, if you square both the sides, the square of it will be sin square x plus cos square x. Okay. Plus two sin x cos x. Yes or no. Which is nothing like but saying that one plus two sin x cos x is T square. Correct. Which means your sin x cos x term could also be written in terms of T only. Right. So not only this term, not only this term is substituted, this term has got substituted in terms of T. So converting the equation in a single variable is always appreciated. Is always favorable for us because it normally gets converted to a single variable, maybe a polynomial equation, and we can always solve it. So in light of this, you know, whatever I have done, I can write the expression as, so this expression could be written as T. Now, see, this term is minus two root two, and this term T square minus one by two equal to zero. In other words, in other words, you have got T minus root two T square minus one equal to zero. Correct. Let's try to simplify this. You can write this as root two T square minus T, minus root two equal to zero. Correct. Yeah. Is it factorizable? Yes. Minus two T plus T. Okay. So take root two T common. So it's T minus root two. Take one common T minus root two. So this will become root two T plus one times T minus root two. Okay. Is it fine? So from here, two values of T come out. One is T is minus half. And other is T is root two. Okay. Let's try to solve it now. By the way, any questions, any concerns so far? Now, many people ask me, sir, why did you give a minus here? Will it work for minus? Yes, why not? Even if you have sin x minus cos x and if you square it, you can still obtain sin x into cos x term from there also. Okay. So it doesn't matter whether it is sin x plus cos x or sin x minus cos x. Take it as a T. Both will help you to get sin x into cos x expression in terms of T only. Okay. So there's nothing to worry even if there is a sin x minus cos x in the expression. Okay. So now the question is just half solved. So your T was nothing but sin x plus cos x. And we have to solve. Okay. I'm so sorry this is root two. Yeah. You have to solve this equation. And here I have to solve this equation. Okay. So I will request you all to work this out and tell me what is the general solution that you're getting for this and what are the general solution that you're getting for this? Okay. Let me call this as A and let me call this as B. So let's solve A first. Everybody work on A. We have already done such questions in the last class and first one. Okay. So what do we do in those cases? We basically converted it to a harmonic form. Okay. Let me call it was the harmonic form. Okay. I don't want to repeat the process. So cos of x minus pi by four is nothing but minus half. Minus half is cos two pi by three. Correct. So the general solution that will come over here is x minus pi by four is two and pi plus minus two pi by three. Correct. In other words, your x is nothing but two and pi plus minus two pi by three plus five by four. Okay. Now, again, this is not a very correct way to write it. So you have to break this up. So it'll be breaking up as two and pi plus two pi by three plus pi by four and two and pi minus two pi by three plus pi by four. Okay. Let's check how much does it give you? This will give you if I'm not mistaken, two and pi plus 11 pi by 12. And this will give you two and pi minus five pi by 12. Correct. Second one. By the way, second one, everybody would realize that this is an extreme value scenario. Okay. So this will happen. Okay. Let me just write it down. So, yeah, so this will happen when cos x minus pi by four is equal to a one, which is actually cos zero. Okay. So x minus pi by four is two and pi. So x is two and pi plus pi by four and belonging to integer. So three solutions will come up from here. One is this. Another one is this. And the third one is this. So three solutions for this equation. Of course, and belonging to integer. Please do not forget writing this. Right. This is okay. So I would consider this to be a typical J advance type of questions. So you may, you may not get such kind of questions in J main exam. Right. But we prepared for such scenarios, you know, I, you know, we cannot blindly say that is not going to come ever in such exams. But yes, it is of a slightly higher, you know, thinking process. So may not be asked in CT or J main J main. It may be. I mean, I'm not very sure, but not in regional interest exams at least. Okay. Is it fine? Any questions? Any questions? Any concerns? Yes. Questions. Your mic was on that. Yeah. Any questions? All right. So I think we have, you know, covered a lot of aspects under equations. So time has now come that we move to trigonometric in equations. Let's try to move to trigonometric in equations to shardly have come into picture. So we are now going to talk about trigonometric in equations, trigonometric in equations. So up till now we were talking about the different types of trigonometric equations and the procedures, the standing operating procedures that we use to solve those type of equations. Now we are going to use more or less the similar strategy for solving trigonometric in equations. Now, is there any standard way to solve trigonometric in equations? Yes. There is a standard way to solve trigonometric in equations also. See, what do we do first? We try to analyze a trigonometric equation in equation in a very small interval, let's say 0 to 2 pi or many times we prefer minus pi to pi also that depends on question to question. I cannot generalize that. So what do we do? We first try to solve a trigonometric in equation in a smaller interval like 0 to 2 pi or minus pi to pi that depends upon the situation. Then what do we do? The upper and the lower limits of that interval whatever we obtain as our answer for that small gap or small interval of 0 to 2 pi. We add n times the periodicity of that trigonometric function involved. Now, it's better to demonstrate it rather than talking about it. Let me demonstrate this with a simple example. Let's say I have, let me take a simple example. Let's say I have sin x greater than equal to half. Okay, let's say I want to solve this equation. Okay. Let's say, let's say I want to solve this equation using my graph. So when you're solving this equation, any question by using a graph, what does it actually mean? You're trying to see. So this is equivalent to trying to say when is the graph of sin x function above the graph of the straight line y equal to half. So for what values of x is the graph of sin x, line above the line y equal to half. Okay. Now, if you just draw the graph of sin x. Okay. This is a sinusoidal function. Okay. It goes on and on, on and on. So no need to worry about a lot of values. Okay. And let's say this is your line, the blue line which I'm making is y equal to half line. Okay. So this is y equal to half line. Right. So this is sin x. And this is y equal to half. Now I want to know all those intervals where your sin x graph is above y equal to half. Right. So what we're going to do is the first going to analyze that function in a very small interval, let's say zero to two pi. Okay. So let us work on zero to two pi. So in zero to two pi, tell me what are the values or what is the interval of X for which the function sin x is above y equal to half. So it says that simple it is in this zone. Correct. In this zone, if you see, I'm just bubbling it up. Okay. In this zone, if you see your sin x graph is above y equal to half line or equal to y equal to half line. Correct. So this is five by six and this is five by six. Correct. If I'm not mistaken. So in that small interval, the answer or the solution for this in equation will be somewhere which is between five by six to five, five by six. Yes or no. Yes or no. Correct. But is this the only possible answer? You'll say no, sir. There will be so many other intervals. Right. It could be this branch also. Right. It could be this branch also and there is no, you know, I mean, there's no limit because this both the graphs are going to extend to from minus infinity to infinity. Right. Okay. So how do we tackle or how do we, you know, take into account all those intervals? So this is what do we do? So what do we do is we just add. Okay. Now tell me what is the periodicity of sin x? What are the periodicity of sin x? Tell me, tell me, tell me, tell me, write down on the chat box. Write down, write down, write down everybody. 2 pi. Yes or no. Right. So what do we do? We add 2n pi to both the upper and the lower limit. Okay. See, this is like, you know, telling the, this is like, you know, telling the examiner that the same interval at a gap of 2 pi will start, you know, giving you the solution, isn't it? So we say union of all these, union of all these values of x. Okay. For n belonging to integer, this will be your answer to this in equation. So this u is basically union, union of all these intervals for n belonging to integer that will become your final answer. Why union? Because even this is your answer. This is your answer. This is your answer. So you have to take a union of all those intervals. So your intervals are like sets, isn't it? Right. So if you take a union of all those answers and start writing it. Is it fine? Any questions? Any questions? Any questions, any concerns? Okay. All right. Let's try one more. Let's try one more. Can you take one more? All right. Let's take Cossacks greater than let's say minus half. Okay. So at this time I would request you to try this out. I'm giving one minute to everybody to try this out. Yes. Should we check this out? Okay. So let us make the graph of Cossacks first. Okay. So this is your graph of Cossacks. I hope everybody knows. Cossacks graph. This is your Cossacks graph. Okay. Now. This question is as good as asking. Which part of or for which values of X is Cossacks graph higher than Y equal to minus half graph. So Y equal to minus half graph is something like this. Okay. This is Y equal to minus half. Fine. Now, all of you please pay attention. So in this case, what I will do as I told you, I can analyze the function either between zero to two pi or between minus pi to pi that depends on the situation. So in this present situation, I'm going to analyze the function between minus pi to pi. Okay. So between minus pi to pi, what is the interval for which your graph is or your Cossacks graph is higher than Y equal to minus half. So you'll say, sir, this is the part where your Cossacks graph is higher than Y equal to minus half. So this interval is minus two pi by three to pi by three. Sorry. Two pi by three. Isn't it? So this for minus pi to pi, this is the answer that you are getting. Now, why do I always choose minus pi to pi? Why do I always choose minus pi to pi? Because you want to account for one complete two pi interval. Are you getting my point? Yes or no? So once I have got this, I can now generalize it. So I can now generalize it by adding the periodicity of cost function. So you have to write two n pi. Just add it to the upper and the lower limit. Like this and take a union of all these answers for all n belonging to integers. So this is the X interval in which, you know, your function, your X should lie, the variable should lie for this inequality to be satisfied. Is it fine? Any questions? Any concerns? Anybody? See, if you take zero to two pi, what will happen is, if you take zero to two pi, then two pi basically starts and over here. Then what do you have to do? You have to break that interval. See, your answer will be like this. So your answer will be zero to two pi by three. And then this value. Okay. This value. This value is finally going to be four pi by three. So from four pi by three to two pi, right? Right? So you have to take a union of those two answers. And then again, you will add to n pi to the upper and the lower limits. And then take a union of the union itself, which is going to make it a little bit complicated. I'm not saying it is going to be wrong. Don't worry. Nobody's going to mark that wrong, but it is just going to make it look complicated. That's it. So then you can follow zero to two pi also. But in that zero to two pi itself, you have to take a union of two answers, right? Correct. And when you take a union of two answers and then again added to n pi and then again take a union, it will make it look complicated. But however, having said that, it all depends upon the option given to you. Okay. Many things depend on the options, especially in this chapter, we, we basically act as per the option. Right. Does that answer your question? Okay. All right. So let's take one or two more questions. We will not go into a lot of questions because, you know, this topic is very straightforward. Yeah. So here in this question, they have thankfully asked us only to find the, the answer between zero to two pi. Okay. Let us also do one thing, not only between zero to two pi, but also in general, we will find out the answer. So find out the general solution also for this, as well as the solution lying between zero to two pi. Let's do both the parts, if at all we can. Is Anita there in the session? Anita. Anita here from this school. Yes. Any success? Anybody? Okay. So this question is basically a simple factorizable question. So you can just break this as minus, minus six sine X minus two sine X plus three less than equal to zero. Take a two sine X common. So it becomes two sine X minus three. Take a minus one common. Two sine X minus three less than equal to zero. Right. So this is as good as saying two sine X minus one times two sine X minus three less than equal to zero. Yes or no. Now remember this fellow is always a negative fellow. Why? It will always be a negative fellow. Why two sine X minus three will always be negative. Can anybody tell me that? Anybody? Simple. Come on. Why two sine X? Right. Because sine X max value is one. So two sine X minus three will always be a negative quantity, which means something into negative is negative. That means that something should be positive. In other words, you're trying to solve this inequality. Yes or no. And this is something which we already solved a little while ago. So your answer will be, your answer will be first of all, in zero to two pi, your answer will be just this. Okay. So if you're looking for a solution. Okay. In your zero to two pi interval, this will be your answer. And if you're looking for a general solution, you just have to add a two N pi to both the upper and the lower limits. Is it fine? Any questions? Any concerns? Okay. Of course we have to write a union here. Sorry about that. Okay. So it's a union of all these intervals. Okay. Any questions? Any concerns? So we are going to now take up the last question on this topic and post that we are going to start with the clinic section ellipse. Right. So ready for the last question? All right. So let's move on to the last question for this topic. Which is let's say. This one. Solve this. Solve this trigonometric in equation. Solve this trigonometric in equation. Yes. Anybody with any success? All right. So let's look into this. So let us bring everything to the left hand side. Okay. So I'll bring the tan square X to the left side. I'll bring a three tan X to the left side. Greater than zero. Okay. Fine. I'm sure it is factorizable. You can just take a tan square X common from this. Take a minus three common from this. Okay. So this is as good as saying tan square X minus three times tan X minus one is greater than zero. Okay. This further is factorizable. I believe tan X minus root three tan X plus root three. Okay. Now all of you please pay attention. I'm using your wavy curve method to solve this inequality as well. Okay. So you can consider tan X to be like a Y. Okay. So let tan X be, you know, why? So it is like solving. You can say polynomial inequality. Okay. You're trying to solve a polynomial inequality. So you can use a wavy curve sign scheme for this as well. So let me just make the roots of these factors. So this is negative root three. One root three. Okay. So plus minus plus minus. Correct. So it is going to be positive only when Y lies between negative root three to one or root three to infinity. Right. But this is your Y value, but Y value is not what I need. I need, I need the value of X. Isn't it? So ask yourself if your tan X is between, so let me make the graph of tan X also for you. So we all know tan X graph is something like this. Okay. And let's say this is your value negative root three, negative root three, and this is that's a positive root three. And this is one, let's say. Okay. Now ask yourself, ask yourself. If your tan X is between negative root three to one, what is the interval of X in which it should happen? Let's say I analyze it between minus pi by two to pi by two only. So your X should lie between all of you please pay attention minus pi by three to pi by four. Yes or no. And for three to infinity, it should lie between pi by three to pi by two. Am I correct? Am I correct? Does this make sense? Yes or no. So what I'm going to do now in this, I'm going to add periodicity of tan X, which is pi into N to all the upper and the lower limits that you see and take a union of that answer. So my answer would be final answer would be X belongs to union of union of N pi minus pi by three to N pi plus pi by four again, there will be a small union coming up over here. In this case, we don't have any option, but we have to write two unions as far as possible try to avoid, but there is no other option left but to write in this case. Okay. Union N pi plus pi by two where N is an integer where N is an integer. Is this fine? So why did I write N pi here? Why not two N pi? Because we are dealing with tan X function, right? We are dealing with tan X function and tan X function is periodic with a period of pi. Okay. Had we, you know, got sin X or cos X function or C kicks for C kicks function, we would have written two N pi, not not N pi. Is this fine? Any questions? Any questions? Any concerns? So with this, we are going to close this particular chapter of trigonometric equations and we are going to start with a new one today, which as I already told you is conic sections. Fine. So before I move on to the next slide, please ask any doubts, any questions, any concerns that you have with respect to this. So you will receive a separate assignment sheet for this topic. Okay. After today's session. Great. So let's move on. Thank you.