 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue solving trigonometric inequalities. They are not very difficult. There are three of them, which I would like to talk about today. So, well, let's just try to do it. I'm sure you tried to solve it themselves first. Do it just by yourself, thinking about these equations. The inequalities, it's very important. So, regardless of the fact whether you have solved them correctly or not, so let's just examine what I consider might be a good solution. All right, the first one. Sine vex plus cosine vex less than square root of 2. Okay. Obviously, we can do it in many different ways. I think what this particular problem actually is trying to make a point, before solving anything straight without any thinking, is to just draw a graph of this function, sine of x, cosine of x, and then together get some curve basically and try to maybe resolve the equation when this is equal to square root of 2. Yes, you can do it this way. But obviously, it would help if some very simple transformation would help you to simplify this particular inequality. So, how to simplify it? Well, in this particular case, I don't know, this square root of 2 actually kind of reminds me something. It reminds me that square root of 2 over 2 is a cosine and a sine of pi over 4. And if I will multiply by square root of 2 over 2 left and right part, it looks like I'm multiplying by sine and cosine, and that would look like a formula for cosine of the difference between x and pi over 4. Right? So, I multiply left and right part by square root of 2 over 2, but since it's the same, so instead of square root of 2 over 2 with the sine of x, I will put sine of pi over 4, and instead of square root of 2 over 2 with a cosine, I will put cosine of pi over 4. And then it will be cosine of x times cosine of pi over 4 plus sine of x and sine of pi over 4, which is the formula for cosine of the difference between x and pi over 4. So, cosine of x minus pi over 4 less than 1, right? Now, cosine is always less than or equal to 1, and at what particular points the cosine is equal to 1, which we have to exclude actually from the solutions. Well, remember the cosine is abscissa, right? So, cosine is abscissa, and the cosine is equal to 0 at the angle equal to 0 plus 2 p m, right? So, x minus pi over 4 should not be equal to 2 pi m, where m is any integer number, because v is 2 pi m is exactly this particular angle, which means that x should not be equal to pi over 4 plus 2 pi m, where m is any integer number and any other x, but this one would satisfy the inequality. That's it. So, again, the trick is to multiply both parts by this expression. Just, you know, think about this. If you have sine and cosine in this particular combination, it might actually be helpful to multiply it by square root of 2 over 2, which gives you a little bit simpler formula, basically. Second one, sine of 2 pi cosine of x greater than 0. Well, this is kind of a complicated thing, so you cannot really, like, immediately come up with the solution, even graphically, quite frankly. I never actually attempted to draw a graph of function like this one. However, we do know how to solve this particular inequality, right? So, we can convert condition on a sine into condition on the y. Well, in this case, y is this. So, when exactly the sine is positive, well, sine is ordinate, right? Remember. So, these angles, they all have positive ordinate. So, it's from 0 to pi plus 2 pi n on both sides. So, these are conditions for the angle sufficient for its sine to be positive. So, this is exactly the condition which we would like to impose on whatever we have under the sine, whatever the argument to a sine is. So, what I want to say is that 2 pi cosine of x should be less than 2 pi n and greater than pi plus 2 pi n, 5 plus 2 pi n. And this is 2 pi n, right? Well, obviously, we should reduce by 2 pi. It's positive, so everything is fine. Over 2 plus n, right? That's what we have. Sorry, that's 1 half. We reduce by 2 pi. We divide it by 2 pi, so it's n. This is the cosine. This is 1 half. And this is n. This is basically a condition equivalent to this one. We did not really lose anything. We did not add anything. So, instead of solving the original equation, we will solve this. And by the way, this should be simultaneous. So, it should be like intersection of two intervals or more intervals with the regions. All right. So, let's talk about this and consider this to be my problem. Now, let's think about it. Cosine is a restricted function. It's from minus 1 to 1. So, n cannot be very big positive or very small negative value because the cosine will be outside of this range. Now, can n be, for instance, minus 2? If this is minus 2, this is minus 1 and a half. Cosine cannot be less than minus 1 and a half, right? Because of this. Now, can n be minus 1? It's minus 1 and this is minus 1 and a half. That's possible. So, one of the things is, where we should look for solution is this. Can n be equal to 0? Well, yes, because it will be from 0 to 1 half, which is also good. From 0 to 1 half, it's possible. Can n be equal to 1? No, because 1 and 1 and a half cosine will not be in between these two values. So, basically, we have reduced our inequality original to two independent solutions. Solutions to this and solution to this. All right? Well, that's kind of easy. So, let's just solve them. For the first one, let's consider the graph. That would be easier. So, this is 0. Pi over 2. This is pi. This is minus pi over 2. And this is minus pi. Now, we also have minus 1 half and 1 half. So, this is 1. 1 and minus 1. 1 half is here. Okay, let's consider the first one. The first one is from minus 1 to minus 1 half. Now, minus 1 to minus 1 half seems to be like this is minus 1 half. So, it's this particular area and this particular area. Right? So, what is the value of X if cosine is equal to minus 1 half? Well, that's pi minus pi over 3, am I right? Now, cosine of pi over 3, which is 60 degrees, is equal to 1 half, right? Cosine of 60 degrees and sine of 30 degrees are 1 half. Now, when I'm changing from pi over 3 to pi minus pi over 3. Now, obviously, if this is pi over 3, this is pi minus pi over 3. And abscissa would be the same absolute value but negative. So, we need minus 1 half. So, this particular point is pi minus pi over 3, which is 2 pi over 3. Now, this point correspondingly is minus 2 pi over 3 because cosine is an even function. And these are obviously minus pi and minus pi. So, the solution to the first one is to make cosine from minus 1 to minus 1 half, we have to make abs from minus pi to minus 2 pi over 3. Looks like I'm right. And from 2 pi over 3 to pi. This is this area. And obviously, I should add 2 pi and 2 both sides. So, these intervals are solutions to the first one. Okay. Now, solutions to the second one from 0 to 1 half. So, from 0 to 1 half. So, it's this. From 0 to 1 half. It's this piece and this piece. The cosine would be from 0 greater than 0 but less than 1 half, right? So, what are these intervals? Well, 0 is obviously minus pi over 2 and minus over 2. And 1 half would be pi over 3 and minus pi over 3, right? Because the cosine of pi over 3 is 1 half. So, in this particular case, we have from minus pi over 2 to pi to minus pi over 3. From minus pi over 2 to minus pi over 3 and from pi over 3 to pi over 2. That's this piece. And again, I should use 2 pi and in both cases. So, these are four different groups of intervals. Now, why groups? Because of this n. n is any integer number. So, for any specific n, we have four intervals. Which is this one, this one, this one, and this one. Where this particular condition is. Where original inequality holds true, right? Okay. That's it. So, what did we do here? We categorized it. We reduced the original quite complicated inequality into a combination of single ones. So, from sine of something, sine of 2 pi cosine x. We have reduced to condition on 2 pi cosine x. And from that, we have reduced the condition on x. So, it's like two steps procedure. And the third problem which I have is cosine of pi xy greater than 0. Okay. Now, the first question you might ask is, hey, we have only one inequality and two different variables. Well, that's right. Which means that solutions are not expressed in some kind of inequality for x. But it should be an inequality on x and y. And how can we express it? I mean, graphically, in the previous cases, we always had some kind of interval. X from this to this or from that to that. That's one dimensional intervals on the line. Well, this obviously is a two dimensional area on the plane, on the coordinate plane. The question is, how does it look? So, my purpose right now is basically express it not only analytically and analytically to express it. It's very easy. The cosine is positive as you understand from here is the graph. So, this is one, this is minus pi over two and this is pi over two. Right? So, cosine goes like this. So, it's positive between minus pi over two and pi over two and plus two pi n are this, right? So, I can very easily express the condition on x and y. Y basically saying that pi x y should be from pi over two plus two pi n to minus pi over two plus two pi n. Right? But what is this? I mean, in case of intervals, you just see, okay, any x which belongs to this interval is a solution. Now, which pairs of x and y belong to this particular solution? Well, for instance, x and y equal to zero does fit, okay, but what else? So, we better express it graphically. So, my point is right now you cannot really express it analytically better than this one. I mean, obviously you can just resolve it for y or for x or whatever. But it does not really give you the feeling of what exactly the solution is. That's why I would like to approach it graphically. All right? So, how can we approach it graphically? I suggest we slightly change this. Well, first of all, we obviously can reduce it by pi. So, I will have x y between minus one half and one half, x y between one half plus n and minus one half plus, no, two n, sorry, two n. Now, for every n, we have its own area of x and y, you see? Let's say for n is equal to zero, we have x y from minus half to half. Now, what is it? Well, let's just think about it. Let's put n is equal to zero first. Now, our x y is from minus one half to one half. Now, how does it look graphically? Let's think about it. Let's build the first one, x y less than one half. Well, let's build x y equals to one half. That probably divides the whole plane into two parts where it's less than one half and where it's equal to one half. And the division point is where it's equal to one, where it's equal, right? So this is hyperbola, right? Because for positive x, we can say y is equal to one half of one x, right? Now, y is equal to one x is just a hyperbola. One half, it's just slightly squeezed down by now. And on the negative side, it's this, right? So these are two pieces of the curve which supposed to divide the whole area into the whole plane into areas where x y greater than one half and less than one half. So in this particular area, let's say take zero, zero. Zero, zero is obviously less than one half. So the whole area which is in between these two is less than one half. Here x y would be greater than one half and here it would go so greater than one half. So that's how we have divided our plane into different areas, right? Now, what about x y greater than minus one half? Again, first of all, let's draw y is equal to minus one half of one x. Now that would be this, right? These two curves represent this particular graph. And now we need again, how does it divide the area? Well, same thing, point zero, zero is greater. So in this particular case, again point zero, zero which is in between these two belongs to the area where x y is greater than one half. And whatever is outside, this is x y would be less than minus one half. This would be x y would be less than minus one half. So what I want to say is that this particular combination of these two inequalities is this diamond kind of shape. Well, I mean it's obviously not a diamond because it's infinite in both cases. In all four sides actually it's infinite. It's asymptotically approaching the axis but it still looks like a diamond basically, right? So that's what it is. This is this particular area which is represented by this particular two inequalities. When x y is greater than minus one half and less than one half. It's this area in between these four lines. Okay, fine. That's done for n is equal to zero. Let's move on. Okay, so far we have this particular piece of the plane as a solution to our initial inequality. Now, let's move on with n. Let's say n is equal to one, all right? So when n is equal to one, I have here two minus, so it's one and a half. So one and one half x y less than x y less than two and one half, right? What is this? Well, again, let's build two curves. First, x y is equal to one and a half and then we will take whatever is greater. Now, what's the difference between this which is actually y is equal to one and a half times one x? What's different from this? This, if you remember, was one half times x, times one over x. And this is one and a half. So it's higher, basically. So if original was a standard hyperbola y is equal to one over x squeezed down by the factor of two, then this one would be stretched out by factor of one and a half. So it would be higher. It's always higher than this one. And what we are looking for is values which are greater than this one, which is this one. Now similarly, with a negative x, it would be this and the area would be this. How about two and a half? It should be less than two and a half. So that would be, the coefficient would be two and a half and would be probably even, I mean, hyperbola even higher than that. But now the condition is less than. So combination of these two would be area in between these two hyperbolas. So this is not a solution. So in addition to this diamond kind of thing, we have a stripe from one and a half xy to two and a half xy. Now obviously if we increase n, we will have, for n is equal to two, we will have what? Three and a half and four and a half, right? So it would be even higher. So again, two stripes will be added. It will be here and here, et cetera. So I will add stripes. Now with a negative n, I will have let's say minus one. If it's minus one, so it would be what? Minus two and a half and this is minus one and a half. Well, obviously it will be stripes on this side. So in addition to this diamond kind of area, which is the result of this thing with n is equal to zero, I will have additional stripes. These stripes would be added as n is equal to one, two, three, four, et cetera, positive. And these stripes would be added as n is equal to minus one, minus two, minus two. So the combination of all of them is this type of a geometrical area, diamond like in the middle, and then stripes around it. But all of these stripes are asymptotically tend to access so they are narrowly, narrowly and narrowly as we go either to the right or to the bottom or to the left or to the up, right? So that's basically how this area which represents the solution to the original inequality looks like. Well, it might be a little complicated, I don't know, but I think this graphical representation really helps you to better feel what exactly all these inequalities are. And don't get scared by the fact that you have two variables and only one inequality. It just means that the result would be a two dimensional area on the plane rather than one dimensional interval or a set of intervals on the line. Well, that's it. I do recommend you to go through these problems again just by yourself. And don't forget that everything is on Unisor.com where the whole educational process actually can be utilized by you. And it has exams for registered students. Your supervisors or your parents can enroll you into different topics, can consider and mark these topics as completed. If the exams which you did, which you took are passed to their satisfaction, etc. So again, try to do it yourself as much as possible all these problems before you listen to lectures like this. And that's it for today. Thank you very much and good luck.