 Hi, I'm Zor. Welcome to InDesert Education. Well, after listening the lecture about what trigonometric equations actually are, well, it's time to solve few problems in this. So we will solve some trigonometric equations. Now, we do know how to solve the basic trigonometric equation, something like this one. We already discussed that during the lecture. So any kind of a trivial manipulation with this particular equation, something like two cosine of 3x plus seven equals to one. I'm not going to do this type of thing because it's really trivial and it's very easily reduced to the basic one. So my purpose right now is to talk about equations which are not trivial. Don't forget that the whole purpose of this course is not to give you certain skills. Okay, this is the case and if you need a situation like this in the future, you have to do that. No, the purpose is to prepare you to unknown, to challenge your mind with something which you just don't know how to do it and you have to think about how to, in this case, solve trigonometric equation. There are no methodology to solve any particular situation to help you to do anything else in the future. No, every problem is separate and you have to really go through many of them to develop the whole repertoire of different approaches, different methodologies, etc. It all accumulates into your mind. So my purpose right now is to present you with a few trigonometric equations which are not that trivial as I was just trying to exemplify. Okay, so I have three different equations for this particular lecture, so let's do it one by one. Okay, the first one is simple. Sine of x plus cosine of x equals a. Well, obviously, a cannot be anything. As we know, the sine and cosine are limited. They're from minus one to one, so a cannot be 25, for instance, obviously. Well, actually, it's much lower the boundaries of the a, which you can derive with. But anyway, how to solve this particular equation? Well, it goes to some property of sine plus cosine and in the corresponding lecture about sum of two angles, I have a lecture dedicated specifically to linear combination of sine and cosine of the same argument x in this particular case. So I will use exactly the same approach, but instead of general approach as in that lecture, I'll use a very concrete example. So what I'm going to do is the following. I will multiply the equation by square root of two over two. That does not change the solutions, the domain of the functions, et cetera. It's a completely equivalent transformation. Now, let's recall that square root of two over two is a cosine of 45 degrees and a sine as well. So sine 45 degrees equal cosine of 45 degrees equals square root of two over two. Remember, the Pythagorean theorem, if this is one, this is one, this is 45 degree, this is 45 degree, and hypotenuse by the Pythagorean theorem is one square plus one square is two, so it's square root of two. So the cosine of 45 degree of this angle, for instance, is equal to one over square root of two, which is the same as square root of two over two, right? And same is sine. It's exactly the same. It's also one opposite calculus over hypotenuse. So we know this. So that's what I'm going to use. And instead of square root of two, in this case, I will put sine of 45 degrees or pi over four. I prefer to do the regions if you're talking about equations. Sine x plus, and this square root over two, I will put as a cosine of p over four. Cosine x is equal to this. Now, what is this? Remember formula for cosine of the difference between two angles. It's cosine by cosine plus sine by sine. So that's exactly where it is. Now, and this is already a basic trigonometric equation which we have considered before during the lecture. So x minus pi over four is equal to plus minus arc cosine of square root of two over two a, from which x is equal to pi over four plus minus arc cosine. So that's the general solution. Actually, to be exact, you have to add 2 pi n in this case and 2 pi n because the cosine is a periodic function. So all solutions are expressed in this particular formula. Well, that's it. That's a relatively simple thing. Now, what kind of a lesson I would like to actually for you to learn from this particular thing? Just looking at that particular equation from the very beginning, it's not obvious, of course, what's the approach how to solve it. But however, this is a guess, if you wish. You really have to be prepared that solving certain trigonometric equations, well, actually solving any good mathematical problem, requires you to find a good approach. Because most likely you never dealt with this particular problem before. So every time you solve another problem which you don't know how to solve, it's very important for you to come up with new methods. That's what develops your creativity. And that's the whole purpose of this course, by the way. So I am trying to present to you as many things which you just don't know what to do. And you have to really think about what to do. And that's why I'm always asking you to try to solve all these problems before you listen to the lecture where I'm explaining the solutions. So by the way, if you didn't do it for this particular lecture, I do encourage you to pause the video and try to solve these two remaining problems just by yourself first, and then listen to the lecture. All right, next one. Next one is sine square x plus sine square 2x equals sine square 3x plus sine square 4x. Again, I basically claim that this is not your typical equation, which you know beforehand, or you are taught beforehand how to solve. You have to invent a new methodology to solve this particular equation. All right. So first of all, if you are thinking about methodology, how to solve this equation, first which I personally think is natural, you should think about these being square of something. Nobody likes squares. Everybody likes the first degree. So the lower degree, the easier it is. Can v represent square sine square x in terms of lower degree of some trigonometric function? Well, yes, of course. Remember this. Cosine 2 phi is equal to its cosine of phi plus phi. So it's a sum of two angles, which means it's cosine phi times cosine phi, which is cosine square minus, this is plus phi plus phi. So this is minus sine phi times sine phi, which is sine square phi. Or since I need sines, I can replace cosine square with 1 minus sine square, because cosine square plus cosine square is equal to 1. So that's 1 minus sine square and another sine square. So it's 2 sine square phi. OK, this is the key which I can use to convert sine square with a clean cosine. So sine square phi equals 1 minus cosine 2 phi over 2, from this. 1 minus cosine sine square goes here and divided by 2. So that's what I'm going to use, where phi can be either x or 2x or 3x or 4x, and all squares will be turned into the first degree of the cosine. All right, fine, let's do it. So what I have is 1 minus, if phi is x, I have 1 minus cosine 2x over 2 plus 1 minus cosine 4x over 2 equals 1 minus cosine 6x over 2 plus 1 minus cosine double 4, which is 8x over 2. That's the equation which I have. Now, obviously, I can multiply it by 2. And 1 will be reduced. It's 2 on this side and 2 on this side. And then minus, minus, minus, minus, I will change to plus, plus, plus, plus. I'll just multiply by minus 1. And the final equation which I have, which seems to be easier than the original, is this one. Now, these are completely equivalent transformations. So it's invariant transformations. So let me wipe out this. Actually, I don't need this one either. So I have reduced my original trigonometric equation to this form using invariant transformations. And this looks easier, right? Now, what's next? So first, I converted the second degree to the first degree. Now, I have pluses and minuses of different cosines. Now, I have it on the left. I have it on the right. And if you remember, any sum of two cosines, or difference between the cosines, can be represented as a product. And that's exactly what I'm going to do. And I would like, if I will convert it into product, something, if I do it smartly, something might actually be reduced. So let me try. Now, instead of 2x, now you remember actually I did exactly the same whenever I was talking about trigonometric series. If you have a cosine of the same of the angles which are separated by the same difference, I divided the difference in half. And I converted every member as half minus or half plus, this particular. So if I have 2x and then 4x, then I can have 3x. And this is 3x minus x. And this is 3x plus x. 2x is 3x minus x and 4x is 3x plus x. So that's what I'm going to do here. And then, if I will do it correctly, I hope it will be, something will be reduced. So this is this. And here I will do exactly the same thing. In between, you have 7x. So it's 7x minus x plus 7x plus. So that's the same equation. I just replaced 2x with 3x minus x and 4x with 3x plus x. And the same there. Why is it easier? Well, think about this way. This is cosine times cosine plus sine times sine. This is cosine times cosine minus sine times sine. And sines will be reduced. So I will have a product of cosine 3x times cosine x. That's it, right? Cosine 3x plus cosine 3x times cosine x plus sine times sine. And this is minus. Actually, there is a multiplier, too. One from here and one from there. And here is the same thing. Cosine 7x times cosine x, then plus sine times sine. And this will be minus sine times sine. So that's exactly what will be left. And obviously, I can reduce by 2. Now, is it easier? Of course it is easier. Because number one, we have common multiplier. So immediately, we have cosine x equals to 0 as potential solutions to our equation. Because if it's equal to 0, then left part is 0, and the right part is 0. Now, this one is x is equal to pi over 2 plus pi n. OK, so we found a few solutions. Now, we want to find other solutions when the cosine of x is not equal to 0. But if it's not equal to 0, I can reduce it. And have only cosine 3x equals to cosine 7x. And then I will do exactly the same. It's cosine 7x minus cosine 3x is equal to 0. And that will convert the difference between these two cosines into a product. So what I will do, the same thing. It's cosine 5x. 5 is in between the 7 and 3, minus, sorry, plus 2x, minus cosine 5x minus 2x, right? Now, this is cosine times cosine minus sine sine. This is cosine cosine plus sine sine, right? So the cosine will be reduced, and the sines will remain. So we'll have from here minus sine 5x sine 2x. And from here, this will be the plus, right? So this minus will be minus again, the same thing. Now, they are identical. So basically, the solution is sine 5x sine 2x is equal to 0. So that's another thing, which has, in turn, two different solutions. Either sine 2x is equal to 0. And the result is 2x is equal to pi n, where n is any integer, and x is equal to pi over 2n. Or sine of 5x is equal to 0, which means 5x is equal to pi n, and x is equal to pi over 5n. So we have three different cases, and this one. But if you look at this, this one includes these points. That's very easy to determine, right? Because this is pi over 2, pi, 3 pi over 2, 2 pi, et cetera. And this one, just every other one from these points. So basically, we can disregard this solution, and these two constitute the solution to a trigonometric equation, which was in the beginning. Which looked kind of scary initially, but after a few transformations, and what we did actually, we always used exactly the same approach. We transformed a difference between trigonometric functions into their product, and then something actually was reduced. That's it. Next one. Next one is the game. It looks a little unusual, but it's still kind of solvable, more or less the same methodology. Sine of pi cosine x is equal to cosine of pi sine x. Yeah, looks, again, a little bit unusual. Scary, if you wish. So what can we do? Well, let's think about what we were using before. We were converging a difference or a sum of cosines or sines into the product. Well, let's try to do exactly the same thing, except we have sine and cosine. They are different functions, and we don't really know what sine minus cosine is. I mean, how to convert it. But I can always convert cosine into a sine, right? Remember, cosine of pi is equal to sine of pi over 2 minus pi. This is one of the classic identities of trigonometry. So I will replace the cosine on the right with the corresponding sine, and then I will bring everything to the left. So we'll have sine of pi cosine x minus sine of pi over 2 minus pi sine of x equals to 0, right? So cosine is equal to sine of pi over 2 minus angle, and then I transferred it to the left. OK, great. Now, let's think about how to convert difference between the sines into their product. And that's very useful, because when it will be the product and it's equal to 0, which means every multiplier is equal to 0 by itself, right? So it's easy. I'll use exactly the same approach as in the previous problem. Remember, if you have sine of alpha, in this case minus sine of beta, how to convert it into product and a product of different trigonometric functions? Well, alpha can be represented as sine of alpha plus beta over 2 minus beta minus alpha over 2, right? Minus beta, yes. And this can be represented as alpha plus beta over 2 plus beta minus alpha over 2. Alpha will be reduced, beta will be 2, yes. And this is the difference between the sine, the difference between two angles is what? Sine cosine minus cosine sine, right? So it would be sine of alpha plus beta over 2 times cosine beta minus alpha over 2 minus cosine alpha plus beta over 2 sine beta minus alpha over 2 minus sine cosine plus cosine sine. But this is minus in front, so I will have everything with a minus sign. Sine alpha plus beta over 2 cosine beta minus alpha over 2 minus cosine alpha plus beta over 2 sine beta minus alpha over 2. So what's reduced? This one. So I will have minus 2 cosine alpha plus beta over 2 sine of beta minus alpha over 2. That's what I have, right? So that's what sine minus sine alpha minus sine beta is. This is how to convert it into a product. OK, so let's use it in this particular case. So this is alpha. This is beta, right? And since it's equal to 0, minus and 2 are insignificant. I will retain only cosine and sine. So it would be cosine of alpha plus beta over 2. So it's pi cosine x over 2 plus pi over 4 minus pi sine x over 2. Right? This is sum of two angles alpha plus beta divided by 2. So it's half of this, half of this, and half of this times sine of the difference beta minus alpha, which is pi over 4 minus pi over 2 sine x minus pi cosine x over 2. And this is equal to 0, which means this can be equal to 0, and that's the solutions. Or this can be equals to 0, which is the solution. Well, let's just consider two cases separately. So when the cosine of something is equal to 0, when it's equal to pi over 2 plus pi n, right? So pi cosine x over 2 plus pi over 4 minus pi sine x over 2 equals to pi over 2 plus pi n. So that's one set of solutions. And as you see, pi is conveniently reduced, right? So if we will multiply everything by what? By 2, right? So we reduce pi, and we multiply by 2. I will have cosine x minus sine x equals 2, pi over 2 minus pi over 4, it's pi over 4. But you multiply by 2, so it's pi over 2 plus 2 pi n. Am I right? I'm sorry. 1 half. There is no pi here. It's 1 half minus 1 quarter, we reduce pi and plus n. OK, so that's one thing. Now, the second thing is when this particular expression is equal to 0, and when sine is equal to 0, when the angle equals to pi n, right? So pi over 4 minus pi sine x over 2 minus pi cosine x over 2 equals to pi n. Again, pi is conveniently reduced. So what do we have instead? We have sine and cosine. Let's put it to the right with a plus sine. So we'll have cosine x plus sine x equals 2. So we multiply by 2, so it's 1 pi over 4, so it's 1 half. Now, pi n would be on this side. So it's n, actually, because pi will be reduced. And we multiply by 2, so it's minus 2n. Now, since n is any integer number, it doesn't really matter whether we put n, minus 2n or plus 2n. So it's exactly the same thing. So we have two different equations, actually. So our one equation, whatever it was in the beginning, is split into two different equations. Well, actually, not two, infinite number of equations, because n can be any integer number, but can it actually not? Let's consider this is cosine and sine. Each one of them is not greater than 1 and not less than minus 1. So the sum of them or the difference between them, just by definition, cannot exceed 2 or be less than minus 2. So obviously, n cannot be anything. n can be 0. That's true. But if n is equal to 1, as you see, in both cases, we have 2 and 1 half. So it's completely out. It's greater than 2. And this cannot exceed 2, because each one cannot exceed 1. Now, on the negative side, minus 1 is still possible, because minus 1 would be what? Minus 3 seconds, right? So we have either 1 half or minus 3 halves. And same here. So that's the only available ends. n is equal to 0 or n is equal to minus 1. When these equations at least make sense. But let's go even further. Now, you remember that the first equation which I did today in the lecture was cosine x plus sine x is equal to a. Remember this? And I solved it. So let's solve it again. So what I did the first time, I multiplied by square root of 2 over 2 and having cosine x cosine pi over 4, that's square root of 2 divided by 2. Plus, well, let's start with the top one, with a minus. Minus sine x sine pi over 2 equals 2. Well, let's do 1 half first. So I multiplied by square root of 2 over 2, right? So that's square root of 2 over 4. That's my equation with 1 half, which is actually what? A cosine of x plus pi over 4 is equal to square root of 2 over 4. From this, we can say that x plus pi over 4 is equal to plus minus arc cosine square root of 2 over 4 plus 2 pi. I'll use another number, k. k is signifying any integer number. So x is equal to minus pi over 4 plus minus arc cosine square root of 2 over 4 plus 2 pi k. So that's what I have if I take 1 half. OK, fine, this is done. Now, what if I have minus 3 halves? If I have minus 3 halves, I will have minus 3 halves times square root of 2 over 2, right? Which is minus 3 square root of 2 over 4. Now, let's think about it. Square root of 2 is approximately 1.4, et cetera. Times 3 would be 4.2, et cetera. Divided by 4 would be greater than 1 by absolute value. Actually, in this case, it's less than minus 1. And the cosine cannot be less than minus 1. So the whole case does not have any solutions, because sine cannot be less than minus 1 or greater than 1. So this is no good. Now, let's go to this one. So it's plus. And again, let's start with 1 half. So I have here square root of 2 over 4. Now, since this is a plus, this is a minus. Cosine of x minus pi over 4 is this. And it's equal to square root of 2 over 4. And now I have minus here. So the x is equal to plus pi over 4. Now, to combine both solutions, the old one was minus, and the new one was plus, I'll put plus minus here. So now I have basically four different combinations of pi over 4 and arc cosine of square root of 2 over 4. Plus plus, plus minus, minus plus, or minus minus. And now, so this is done. Now, for minus 3 half, I have exactly the same consideration. I will have minus 3 quarters of square root of 2, which is by absolute value greater than 1. So there is no solution. So basically, that's it. We finally got all the solutions. And that's the result, where k is any integer number. Well, that's it for today. These three problems are important in some way, because I was trying to put something non-trivial. But let's just think about what kind of methodology I was using. Primarily, it's just converting minus sine or sine, plus sine into product. And then the solution is easier, because if the product of two variables is equal to 0, then either one variable or another is equal to 0. So just remember, this is approach. This is the methodology. And whenever something else comes up, this is one of your instruments in your toolbox. And that's the purpose, to put not the exact solutions to all the problems that you can just see in the future, but to equip you with sufficient number of tools that you can use to achieve the goal. Well, that's it for today. Thank you very much. And good luck with new problems, which I am definitely going to put on my website. Thank you.