 Hi, I'm Zor. Welcome to Unisor Education. I would like to start talking about trigonometric equations. So this is an introductory lecture and obviously it will be followed by problem-solving lectures. Alright, so trigonometric equations. Probably you can say that any equation where one of the components is a trigonometric function is a trigonometric equation. There are some simple ones, like, for instance, 2 sin of x equals to 1. Well, some obvious solution is, well, then sin of x is equal to 1 half, and we do remember that the sin of 30 degrees is 1 half, so it's probably x equals to 30 degrees. Now, is that it? Is that the solution? Well, obviously not, because number one, sin is a periodic function. So whenever you have some value for x, obviously the value which is increased by 360 degrees will also be a solution. So 390 would be a solution or anything else. Also, not only for an angle equals to 30 degrees, sin is equal to 1 half. Also, if I'm not mistaken, it's 150, and that's also 1 half. So it's not that simple, it's still simple, but it does require certain accuracy, I would say. And then there are obviously some much more complicated equations which also can be called trigonometric equations. I don't know, something like square root of sin x plus log times 2 of sin x plus 5 equals to 0. I don't know, I just write something which doesn't have much sense. It looks very complicated and quite frankly I don't even know how to solve this. Maybe if I really try, maybe I would. But that's not really what I would like to present you right now. I would like to present you with simple trigonometric equations which we can solve analytically to come up with some formula actually, which is the solution to these trigonometric equations. And then gradually increase the complexity of this. So forget about this, we will never be talking about this nonsense. So let's start with simple trigonometric equations. And the first one which I would like to talk about is the following one. Don't be surprised that this is tangent. For some reason traditionally we used to think that sine and cosine are the first trigonometric functions we are talking about. And then we go to tangent because actually a tangent is defined as sine over cosine. But in this particular case, in the case of equations, it's easier to start with tangent and you will understand why a little later. Now, if you would like to solve an equation there are many different approaches obviously. You can do it analytically, you can do it graphically, you can do it numerically, etc. What I would like to emphasize is the graphical aspect of this equation and some others as well. Now, if you have a general equation which looks like that, where f is some function and you would like to address this particular equation from the graphical standpoint, how would you approach it? Well, you would do the following. You would first draw the graph of this function which is something, whatever it is. Then you would draw the graph of this function which is just a straight line parallel to the x-axis at point of y equals to 8. So it's this. And wherever these two are intersect, intersecting each other. So the graph of the function f of x and the straight line which is a graph of this function. So it's this point, this point, this point, everywhere else wherever it exists. Take their x-coordinates and these are solutions because at point x1, f of x1 is equal to this, according to this graph. And it's a as well. So that's why f of x1 is equal to 8. Same thing f of x2. It's this point and it's also 8. The origin of this point is 8. And so is f of x3. So x1, x2, and x3 and any other abstice of intersecting, of intersection of these two, graph and straight line. These are all solutions. Okay, we know that, so let's go to tangent. This is just a general theory about graphical solutions. What's the graph of the tangent? Well, first of all, if you don't remember it, I do suggest you to just stop this lecture and go to the tangent lecture and just refresh your memory. I'll just tell you the following that if this is a tangent, then the graph looks like this. These are vertical lines and tangent, this is minus p over 2, this is p over 2, this is 0. And tangent is asymptotically close to these vertical lines on this particular interval from minus p over 2 to p over 2. It's not defined at the end of this interval, but it's defined everywhere in between and it takes all the values from minus infinity to plus infinity. And that's the graph of this function, tangent of x on this particular period. And then this period actually repeats itself. So let me just have one more thing. So it's about this. So this is pi. And same thing here, this is minus pi. So that's the graph of the function y is equal to tangent x. Now, this is the line y is equal to a. So this is a. So where are our solutions that are here? So if you drop down the perpendicular from these points where the line y is equal to a intersects with the graph of the function y is equal to tangent x. These are x where tangent is equal to a. And these are solutions. Now, obviously, since the tangent is a monotonic function on the interval from minus pi over 2 to pi over 2, and it takes all the values from minus infinity to plus infinity. Obviously, there is one and only one point within this interval where the tangent is equal to one concrete value a. And what is this point? Well, if you remember, and again if you don't go to the corresponding lecture about inverse trigonometric function, this is actually a definition of the function arc tangent of a. So the function arc tangent of a is inverse to tangent, but only on this particular interval from minus pi over 2 to pi over 2. While argument of the function arc tangent is changing from minus infinity to plus infinity, the result of this function is from minus pi over 2 to pi over 2, not including the ends of this interval. So this is a solution to this equation. Now, as we know, this particular interval, which is a base interval of monotonic behavior, is actually also a period of the function. So it exactly repeats left and right up to infinity, which means that we can add to this value a period, a period is equal to pi, or two periods, or three periods, or subtract a period, or two periods, or three periods, which means that the total solution, the set of all the solutions of this particular equation can be written as x is equal to arc tangent of a plus n pi. Pi is a period, let me just get rid of this, arc tangent of a. So if a is this constant in the equation tangent x is equal to a, then the solution to this equation x is equal to a solution of the equation in the base interval where the function is monotonic. And this is the period plus period times any integer m, plus or minus, basically integer can be positive or negative. So the set of infinite set of solutions, because n can be any integer number, countable obviously, because integers are countable. So the solutions, there are infinite number of solutions to this particular equation, and these are described by this particular formula where m is any integer number. That's it. No more complications for tangent. So we have actually solved the equation in two steps. First, we determined the base interval where the function is monotonic, and that's why we can very easily invert this particular equation and determine the argument if we know the function. Now, if the function is not monotonic, like parabola for instance, and you have one particular value which you'd like to find where are the arguments where the parabola takes this way, there will be two. But in this case, if the function is monotonic, it's always one and it's determined on this particular interval of base interval of monotonic behavior. It's determined by this particular formula, and this is the definition of the function arctangent, so I don't have to really explain why it's that. And then we just add the period. And again, let me just emphasize one more thing. The base interval of monotonic behavior and the period of the function are exactly the same. That helps in this particular case, which will not be the case for sine and cosine, and that's why I started with tangent because it looks simple. Now let's go to cotangent, and that's as simple actually. We just have to use different intervals, but it's also the same type of solution. So we have equation cotangent of x equals to a. So back to the graphic of cotangent. Cotangent is slightly different, but again, if you don't remember it exactly, go to the corresponding lecture for cotangent and this graph, and you will see that the period is also pi. However, the base interval of monotonic behavior is not from minus pi to plus pi like in tangent, but from zero to pi. That's the graph. So the base interval is, which we are considering is from zero to pi, not including the nth, of course, where cotangent is not defined. The graph looks like this. It's monotonic. It's monotonically decreasing, not increasing like tangent, but what matters is that it's monotonic. Which means this equation always have one and only one solution, which by definition of the function arc cotangent. So by definition of the function arc cotangent, the arc cotangent of a is a solution to this particular equation in this particular interval of monotonic behavior, which happened to be at the same time the interval which is the period of the function. So all we have to do now is to add the period. So if you have here 2 pi and it goes like this, so you have another value, they are different by pi or 2 pi or 3 pi or minus 2 pi. So the whole solution will be arc cotangent of a plus n pi. Now, obviously I'm not actually talking about some peculiarities like for instance the cotangent is not defined at zero or pi or 2 pi etc. But these situations would not occur because the arc cotangent of a always exists, doesn't matter which a is, wherever a is there is always the intersection. And it's always in between zero and pi not including the n's. So I don't really have to think about can I have a solution x is equal to zero or x is equal to pi. No, I will not have it because of this. Arc cotangent cannot have a value of zero or pi or anything like that. And that's the same thing as before, I actually should have mentioned it for arc cotangent. These peculiar values where the cotangent doesn't exist would not be solutions. This formula doesn't fill out there for this. Okay, fine. These are two simple cases tangent and cotangent. Now let's go to sine and cosine which are just a little bit more difficult. And we will understand why in a second. So we'll go with sine. Let's say sine of x is equal to a. That's the equation. Now again the graph of sine. So this is pi. This is minus pi. This is minus pi over two. This is pi over two. The function goes like this. This is one. This is minus one. This is a period from minus pi to pi. I mean I can take any interval which has the lengths of two pi because two pi is a period. But this is the most convenient interval. I'll explain why. First of all would be observed here that on this period function is not monotonic. Like period for a tangent was pi and the period and the basic interval of monotonic behavior is also has a length of pi from minus pi over two to pi over two. Similar to cotangent. This is not the case. So if I will draw a line y is equal to a, well number one that line might not even intersect the graph. If a is above one or below minus one we don't have any solutions. So the first thing which comes to mind is that the a should be from minus one to one if we want to have any solutions. Because if it's not then there is no solutions. Sine will not be greater than one or less than minus one. Okay so that's one thing. But even after that restriction we still have a problem. You see we have two points where the graph of sine intersects with a straight line. And we somehow have to deal with this. Now in the previous case with tangent and cotangent we had the solution obtained in two steps. One we solved it on a basis, on a base interval of monotonic behavior. Which happens to be the same as the interval of periodicity. And so all we had to do on the second step is just to add period times some integer number n plus n times pi. In this particular case the period has a length of 2pi from minus pi to pi I chose. But the monotonic behavior is on a narrower interval from minus pi over 2 to pi over 2. And that's where our arc sine function is defined. So only on this particular interval from minus pi over 2 to pi over 2 we can really use arc sine. And this particular value, this particular value, this x. This x is in this particular case is arc sine of A. But as we know by definition of the arc sine x is only from minus pi over 2 to pi over 2. What about this guy? This is also obviously a solution. Now here you have to, just from looking at this particular graph you see that this interval is equal to this interval. Now what does it mean? Remember sine of x is equal to sine of pi minus x, right? Remember this? Again if you don't remember this elementary identity check with the corresponding lecture on trigonometric identity. And it's very easy actually to see from the unit circle because what is a sine on the unit circle? It's ordinate, right? So if you have a point here and point here both ordinance are the same. So sine of this angle and sine of this angle are the same. Now if this is x, this is obviously pi minus x, right? So that's exactly the same thing. So using this particular identity I can see that at least for positive A, let's just consider positive A when the intersection is in the interval from 0 to pi. This point of intersection is arc sine. And this one is pi minus the first one, right? So we have two solutions. Pi minus arc sine of A. So this is x1, this is x2. So we have two solutions. This is for positive A. Okay, how about for negative A? Here is A. Now this again, this x1 is arc sine because, so again we have x1 equals to arc sine, okay? And how about this one? Well, same thing. But we have to start from minus pi and add this arc sine. But it's negative. So it's again minus x2 is equal to arc sine is negative, right? It's negative. So if I want to go to the right I have to basically subtract the negative. So it's minus pi minus arc sine of A. These are two solutions. And finally, if A is equal to 0, you have 1, 2 and 3 solutions. Well, let's just not consider this A is equal to 0 right now. We will consider it a little bit later when I combine everything into one formula. Alright, so we have these two cases. Now, how to generalize these to the entire real values of x? Well, let's forget about this graph. We don't want it anymore. So we can say that in this particular case I have to add basically a period to pi. So what happens is x1 is equal to arc sine of A plus 2 pi m. And x2 is equal to arc sine A, sorry, it was a minus sign. Minus arc sine A plus 2 pi m plus pi, right? Pi, this is the pi minus arc sine and plus the pi. That's for A greater than equal, greater equal to 0. Let me rewrite it slightly differently. Minus arc sine of A plus pi times 2n plus 1. I just factored out pi. And this I will not change, I will just rewrite it in the same formula. Pi times 2n. So when n is even, I have plus, not n, sorry, when multiplier by pi, this multiplier, if it's even, then this is the plus. If it's odd, this is minus. I can combine these two formulas into one, which is x is equal to minus 1 to the power of k arc sine of A plus pi k, where k is any integer number. So if k is even number, so it can be represented like this. Then it would be minus 1 to the even power, and minus 1 and even power gives you 1. And if k is odd number, minus 1 to the odd power, like 3, 5, etc., will always be minus 1, so it will be minus sine. So this is just a shorter version of these two formulas for A greater than 0. Now, let's do something similar with A less than 0. So I have to add 2 pi n in both cases. So x1 is equal to arc sine of A plus pi times 2n. And the second one is minus arc sine of A plus pi and plus pi times 2n. So it's the same thing as in this case. So that's minus in this case. It's minus pi. So that's why it's 2n minus pi. But consider the n is any integer number. So 2n is any even number. And 2n minus 1 is any odd number. It doesn't really matter that this is 2n plus 1 and this is 2n minus 1. Still, this actually goes through all the odd numbers and this goes through all the odd numbers if n is any integer. So it doesn't really matter whether it's minus 1 or plus 1. It's still an odd number. Any odd number, as n is moving across the all integers, this is moving across all the odd integer numbers. Which means that I can combine them into exactly the same formula where k is any integer number. So the solution is exactly the same. I don't have to differentiate it anymore. So I can say that this is one and only solution which combines all the values which combines all other solutions in one formula where k is moving along the old integer values. Now, by the way, I didn't want to consider the case a is equal to 0, not because it's any special. It's exactly here. I'm just saying that if a is equal to 0, then it doesn't really matter what k is, odd or even. So arc sine of 0 is equal to 0 because that's the only angle sine of which is equal to 0 in the base interval of monotonic behavior. So it doesn't matter what kind of an exponent here is. It will be 0 and that's why the solution will be only pi k. So for a is equal to 0, the solution is pi times k where k is any integer. Just a simpler formula. I didn't want to put so much attention to this. Alright, so basically that's it for a sine. As you see, it's slightly more complex than this tangent. This tangent within two steps. One is base interval of monotonic behavior, which exactly the same as the period. In case of a sine, we have to do it in three steps. First, the base interval of monotonic behavior where we can use arc sine, the inverse function. Then we have expanded it to a period by just adding, by changing like pi minus arc sine or minus pi minus arc sine. And then the third step is exactly the same as before. The last step is expand using the periodicity. So just one more steps. Two steps for tangent and three steps for sine. And cosine will be exactly the same thing. Let me just do it quickly. So the equation is cosine of x is equal to a. Obviously solution exists only in case of a is in between 1 and 0. So now let's go back to the graph and establish the period and the base monotonic behavior. Now cosine looks like this. From 0 to 2 pi. This is pi. This is pi over 2. This is 3 pi over 2. And the monotonic behavior is from 0 to pi. Here the function is monotonically decreasing. Which means we can solve this particular equation on this interval without any problems. And the solution is arc cosine of a. So that's the solution. Arc cosine of a. But there is another one which is here. How can we find this solution? Well, very similarly. Obviously this is equal to this. So cosine, as we know, is an even function. Which means cosine of x is equal to cosine of minus x. And cosine of minus x, since cosine is a periodic function, is equal to cosine of 2 pi minus x. I just added 2 pi to the argument, right? And this is exactly what it is. 2 pi minus this. This is just basically an explanation why if this is a solution, then this 2 pi minus arc cosine of a or minus arc cosine of a are solutions as well. And these are just different in 2 pi. And this is different from this by the sine. And the cosine is an even function. So these are all solutions. Now, if a is positive and if a is negative, it's actually the same thing. Then this would be the arc sine. And again, this length is equal to this length. So basically I don't even have to consider two different cases when a is positive or a is negative. The solutions are always this and this. Now, 2 pi minus would actually fall within the period from 0 to 2 pi. That's why I might actually prefer that to write in this particular case. But it doesn't really matter in the general case when I expand it to the whole set of different areas of arguments because if I add the periodicity, if I add 2 pi n basically, then whether I add this 2 pi here or it will be just part of this 2 pi n doesn't really matter. So the solution are plus minus arc cosine a plus 2 pi n where n is any integer number. That's the general solution to this particular equation. Well, that's it. This is just an introduction to trigonometric equations. Now, the real problems related to trigonometric equations are obviously much more complicated than these and we will consider them in the future lectures. But what I would like to say is that most likely these trigonometric equations which are presented to you as problems will be reduced eventually to these using just regular algebraic methods. Just let me give you an example. What if you have something like sine square of x is equal to one half? Well, it's a combination of two things. Algebraic and trigonometric. Algebraic is sine square. So what if you have an equation which is like y square is equal to one half? How to solve this? Well, our business solutions are y is equal to plus or minus 1 over square root of 2, right? Or plus or minus square root of 2 over 2, which is the same thing. And now we have to think about, okay, y is not just y, this is a sine of x. So I have to solve the equations sine of x is equal plus or minus square root of 2 over 2, right? And this has its own solutions. For plus square root over 2, it's in the base interval of periodicity. It's 45 degrees. It's pi over 4. And then you have to expand it using the general formula. And for minus, it would be minus pi over 2 and also expand it to the period and then to the entire set of values. So it's, again, a combination. First you do something algebraically until you solve it to the level of one single equation of this type. And then you apply trigonometry to get the final solution for x. That's the general idea. I mean, obviously there are some variations. But most of the equations which are related to trigonometry, which need to be solved, they are of that type. So first you apply some algebraic methodology and then finally you go through the trigonometric part, trigonometric steps, if you wish. Alright, so be prepared that next lectures will be evolved around solving the problems. And that's it for today. Thank you very much for your attention and good luck.