 Hi, I'm Zor. Welcome to InDesert Education. I would like to start a new topic. It's about systems of trigonometric equations. Well, first of all, about the concept. What are the systems of trigonometric equations? Well, you know that trigonometric equation is equation which has some trigonometric functions involved. So systems of trigonometric equations are exactly similar. It's some systems of equations with more than one unknown, and certain trigonometric functions are involved in this system. Now, is there any general methodology of solving the systems of trigonometric equations? Well, no. But regularly, the substitution method, which is similar to substitution method in other systems, in algebraic systems, is usually something which might be involved. Not necessarily. Actually, the more interesting examples of systems of trigonometric equations do contain the substitution as part of it, but that's the most trivial part. The most interesting part starts after the substitution is done, and you have to basically solve certain trigonometric equations. So the best way to learn about systems of trigonometric equations is to solve them. You know, my recipe for all kinds of learning methods is just solve problems. The more you do, the more difficult the problems are. The more your mind is prepared for an unknown, whatever you can meet in your real practical life, and that's the only recipe which I have in this particular case. So no particular skills will be developed by solving trigonometric equations or systems of trigonometric equations. It's just development of your mind, your creativity, your critical abilities, your analytical abilities, your ingenuity and intelligence. All right. So I would like to present one particular system and I will just solve it. Then there will be certain problems which I will just continue analyzing, can explain, etc. So this particular lecture will be about one particular problem, one particular system of trigonometric equations, and let's just follow the logic of this. The system is as follows. A sine x plus d plus sine y equals c and x plus y equals d. So two trigonometric equations with two variables x and y, and as you see, there are some trigonometric functions involved, which allows us to basically say that, okay, this is a system of trigonometric equations. All the only one of them is really trigonometric and another is more traditional algebraic. Now, this is one of the simplest examples, and it's basically solved using the regular substitution method which you might imagine would do in this particular case. So what do we do? We resolve the second equation for y in terms of x, which is this, and substitute into the top equation to get one trigonometric equation with one unknown x. After we define x, then we can just define y as well. So what kind of trigonometric equation will be as a result? A sine of x plus b cosine of d minus x equals c. Now, obviously, I cannot solve it without certain manipulations, and the first manipulation which basically is natural is to open up cosine of the difference between two angles. So it will be a sine x plus b cosine d cosine x plus b sine d sine x equals c. All right? Cosine of the difference is cosine cosine plus sine sine, and the coefficient b goes to both sides. Now, if you remember in one of the lectures which was dedicated to trigonometric equations, I was solving equation of p sine x plus q cosine x equals whatever a. The way how I solved this equation was I multiply and divide this by r equals to square root of p square plus q square. So I have r p over r sine x plus q over r cosine x equals a, and then I basically interpreted p over r as sine of some angle, and q over r as a cosine of the same angle, and using this, it was possible. So let me remind you the whole technique actually in details, and then I will apply it to this guy. But first we have to convert it into this form. So we bring together everything about sine x and about cosine x. Right? So it's b cosine b cosine x plus a plus b sine d sine x equals c. So basically I will call this p and this q. These are constants, right? Because d is a constant, and a is a constant, and b is a constant. So they are known constants. So instead of going into large formulas and just use the letters p and q, p signifies b cos sine d and q signifies a plus b sine d. So what I have is p cosine x plus q sine x equals c. How to solve this equation? As I was saying before, I will have r equals to square root of p square plus q square. So if I'm talking about solving this equation, I presume that they are not at the same time equal to 0, because otherwise it's not an equation. So that's why r is not equal to 0, obviously. Now p over r can be a cosine of some angle phi, and q over r can be a sine of this angle phi. How can I find this angle phi? Well, let's think about this way. If both p and q are positive, now r is always positive, so cosine and sine are positive. Now cosine, as you remember, is abscissa of this point. This is cosine, and this is an ordinate is a sine. If this is a unit circle, and this is an angle phi. So if p and q are positive, then I can actually say that the phi can be arc cosine of p over r. Actually, let me just invent one particular angle alpha, which is arc sine of absolute value of p over r. So absolute value is always positive, so I always have this angle alpha. So it's not phi, it's alpha. Now angle phi is defined by these identities. So now we have different cases. If p greater or equal to 0 and q greater or equal to 0, then my angle phi is equal to exactly alpha, because it's in the first quarter. Now if p, which is abscissa, is greater or equal to 0, but q is less than 0, so it's here. So abscissa p is greater than 0, and ordinate is less than 0. Then phi is equal to minus alpha, right? This is alpha. This is alpha, this is minus alpha. Now if p is less than 0 and q is greater than or equal to 0, p is here, so it's here. Angle phi is equal to pi minus alpha. And finally, if both of them are negative, then phi is equal to, it's here, pi plus alpha, right? So if alpha is defined as this, where absolute value of p is always positive, that's why alpha is here. Then whatever my p and q are, I can always find angle phi using the angle alpha and pi in this particular case, depending on p and q. And definitely sine and cosine of phi would be whatever I'm looking for. Now, having that angle phi, which is calculated as through alpha and these inequalities, I can say that here, instead of this particular equation, I can put cosine phi, cosine x, plus sine phi, sine x equals c. Because p over r is cosine phi and q over r is sine phi, right? So r is a factor multiplier factor doubt, from which we can always see that cosine of x minus phi, because that's exactly what's in these parentheses, is equal to c over r, which brings us to solution. X is equal to phi plus minus r cosine of c over r plus 2 pi n, right? So x minus phi is equal to r cosine plus minus r cosine of c over r and plus 2 pi n. And then phi should be n. Basically, that's the solution. That's x and then the d minus x would be y, obviously. So y is equal to d minus phi. And since it's plus or minus, it doesn't really matter what I in what order I put it. And n is an integer, so again, it doesn't really matter whether it's plus or minus. So any pair of x and y, which is calculated, actually, you know what? I don't even have to use the same letter n. I can use letter n and some calc, because it's still a period anyway, right? All right. Oh, no, no. And it's important here. And in this case, I should really put minus, because it's minus. So x plus y would always be d. I really have to preserve the sign, so it's not plus minus, it's minus plus, right? Because it's d minus x, so I have to reverse all the signs. Minus phi, instead of plus minus, I should put minus plus, and instead of this plus minus, where n is any integer number, obviously. OK, now x plus y would be equal to d. But as far as this relationship, it doesn't really matter about 2n, 2 pi n, because it's a period of correction. All right, so that's the solution. What's interesting is it's relatively simple system, because all I did was I substituted the unknown to the first equation, reducing the system of two equations with two unknown to one equation with one unknown. And that letter was actually kind of the thing which we used to solve before. It's this type, cosine and sine with some coefficients. I will probably introduce to you some more special techniques whenever we will go through problems related to systems of equations. And again, the more problems we solve, the more these special techniques you will absorb. And that's not, it's just the tools. You don't have to really remember it by Carter and anything like that. But if you try to do it yourself first before listening to my explanations, it will definitely enrich your repertoire of your techniques which you can use. So I definitely suggest you to go again maybe through this example as well just by yourself. And then for whatever problems will be introduced in this topic, try to solve it themselves and then listen to the lectures. Well, that's it for today. Don't forget that Unizor.com contains not only these lectures or problems, et cetera, but an entire educational process which you can actually incorporate into your studying because you can enroll or your supervisor can enroll you. Your parent can enroll you in any topic. There are or there will be exams for each topic. Some of them are not yet ready, but most of them actually are ready right now. And so the exam will give you a very good picture of how well you master the particular topic, particular part of this course of mathematics. Well, that's it. Thank you very much. And good luck.