 Hi, I'm Zor. Welcome to Indizor Education. Well, it's time to solve a few trigonometric inequalities. I did have some introductory lecture about basic trigonometric inequalities like sine effects less than a. So now let's just solve some problems using certain methodology which might really be very helpful. So I have three problems. They're not really difficult, but there are certain calculations which I hope I will not get lost in. Okay, the problem number one, tangent of x plus cotangent of x less than or equal to a. Alright, so the first complication which strikes our eye is that it's not just simple tangent less than something. It's some of two different trigonometric equations. Some of them, each part we can actually solve. Tangent is less than something and cotangent is less than something. That's okay. How about their sum? And what kind of a function is the sum of tangent and cotangent? Just to better understand what this is all about, I prefer to draw a graph. So let me just think about the graph of this particular function. So this is, let's say from minus pi over 2 to pi over 2. I know the tangent is monotonic and it looks like this. That's our tangent. Now the cotangent is monotonic on 0 to pi, right? So it's like this and it's decreasing. And similarly, here we will have minus pi, so it's this, right? Now, I would like to have some common period of periodicity, interval of periodicity. So I will choose from minus pi over 2 to pi over 2. So let me just wipe this out so it doesn't disturb us and this. And let's sum them together. Now, obviously the minus pi over 2 and 0 and pi over 2 are asymptotes because in this point tangent is going to infinity. At this point cotangent goes to sum infinity, whatever it is. And at this point, again, tangent is equal to infinity. So these are asymptotes minus pi over 2, 0 and pi over 2. Now, at this point, they're both of them the same. And what happens then? Then one function goes to infinity and another goes to 0. So basically we are adding something to this and it will go to this way. And on this side, from this point, it will be exactly the same, but it's the tangent which will go to infinity. So the function goes to this. So my total graph would look like this in this case. This is my sum. Now similarly, on the left from the 0, this both are negative. So the sum will be like this. And then it goes to negative infinity. So my graph would be like this. Now, these are points. They are right in the middle, by the way, between 0 and pi over 2. It's pi over 4, actually, which is 45 degrees and cotangent are equal to 1. So their sum is equal to 2. So this looks like this is 2 and this is minus the corresponding one. But this will make sure actually this is really true. But in any case, I wanted to have a feeling of what actually the graph of this function represents. Now, if I would like to have less than or equal to a, well, it depends. If a is between these two, then the bottom part will be completely inside this interval. If it will be below 2, I will have two different intervals, left and right, and it will be above 2, I mean below minus 2. But if it will be above plus 2, I will have the whole branch of these guys and piece of this plus piece of that on the other side of the intersection. So it all depends on where y is equal to a straight line intersects this graph. So in this case, like I just drew, it's greater than 2. Now, all this part would be part of this solution, well, except 0, of course, where it's not defined. And on the right, from the right part, it would be on this side. Again, not including the pi over 2. So that's basically my general approach to solving this particular inequality. Now let's do it analytically. Now, again, we have the problem, tangent and cotangent, and we don't like it. However, we do know that cotangent is 1 over tangent, right? So what I have here is tangent of x plus 1 over tangent of x less than or equal to a. Now, that's a little bit easier, but not that easy yet. Now, obviously, I have to exclude from all the solutions where the tangent is undefined or the tangent is equal to 0 because the cotangent is undefined. So basically it goes to x not equal to 0, pi over 2, pi minus pi over 2, et cetera. So basically we can say that it's pi over 2 times n, where n is any integer number. So these are prohibited values, and we should really exclude them from all the solutions. Now, after this is excluded, I can multiply by a tangent of x because it's not equal to 0 now since we excluded these values. And what I will have? I will have tangent squared x plus 1 less than a tangent of x, right? Or tangent squared of x minus a tangent of x plus 1 less than or equal to 0. Let's call tangent of x by some letter b, some variable. Now, this is, in terms of v, this is a quadratic equation, a quadratic inequality. And we know how to solve this quadratic inequality because, practically speaking, any quadratic inequality can be represented as a parabola. In this case, it's this way. So depending on where exactly are solutions to the corresponding inequality, where the whole left part is equal to 0, solution to this are all these values of argument in between these two points, right? Because that's where the whole parabola goes below 0. So let's solve this equation. We will find what these are representing the solutions to the corresponding equation. So if I have equations equal to 0, v squared minus a v plus 1 is equal to 0, and solutions are 2 a plus minus square root a square minus 4, right? That's what v first and second are. That's the solutions. And our condition when it's less than or equal to 0, it actually means from that with a minus to that with a plus. So in between these two values, my quadratic polynomial is less than 0, right? So we have solved the corresponding equation, find solutions, and in between these two, since the coefficient of v squared is 1, so the parabola goes upwards, and in between these two solutions, it's below the 0 level. So what we have actually reduced our original inequality is that tangent of x should be basically greater than this particular expression, but less than this particular expression. So we have reduced one particular trigonometric inequality, reduced to a combination of two, to the union of these two conditions, right? Well, that's easy because we know what to do in this case because we can solve each one of them and then take the union. What I would like to add, it's very clearly visible here that a should be at least 2 by modules. It should be either greater than 2 or less than minus 2 because otherwise my expression under the square root would be negative and there would be no solutions. So solutions exist only if a is greater than 2 or less than minus 2, which basically corresponds to whatever I was just talking about, right? Okay, now how to solve these guys, okay? If I would like to solve this one, the left part, so tangent is greater than some value. Now, the tangent is this. So if somewhere is wherever it is, it's greater than, so it should be greater than arc tangent, right? So solution to the left part is x is greater or equal than arc tangent. This is minus. That's solution to this one. Now, solution to this one is x is less than or equal to arc tangent of a plus minus 4 divided by 2. Now, and obviously in both cases we should not really go beyond our interval where we are looking for solutions. So if x is greater than this one, then it should be less than or equal to pi over 2 and in this case it should be greater or minus pi over 2, right? So whenever a is greater than 2 by absolute value, we basically have this particular type of equations of inequalities and we have to combine together into some kind of a solution, right? So what we can say is the following. If a is somewhere here, then above 2, then we can say the following. These two points represent arc tangent of a minus square root of a square minus 4. Let's call it p and q. This is p and q. So this is, it's not really p, it's arc tangent of p and arc tangent of q, right? So in this particular case, so what we have is our solutions are all these guys, all the negative except minus pi over 2 and minus and except 0 because that's where it's not defined. And then from 0 to arc tangent of this particular expression of p. So let's call it p. So we will have from minus pi over 2 x to 0 except we don't have it both obviously. And then we have from 0 to including arc tangent of this expression, which I call p, but anyway it's a minus square root of a square minus 4 over 2. And in addition to that, we have another interval from arc tangent of a plus square root of a square minus 4 over 2 to pi over 2. So that's what we have, right? Completely this thing, that's this. And this is, oh, no, no. It's not before that, it's in between these things, so I'm wrong. So we have to combine these two together. So it's not to the left and to the right, it's in between, actually. So it should be from minus x to plus arc tangent of a plus square root of a square minus 4 over 2. We have to combine these two together, right? So that's what we have. So this is exactly the same as whatever I put here except I applied these particular restrictions. No pi over 2 and no 0. That's why we did this. Now, if it's less than minus 2, I will have correspondingly from minus pi over 2. So this is for a is greater than or equal to 2. Now, if I have a is less than or equal to minus 2, I will have two pieces from here. The one from minus pi over 2 to arc tangent. So minus pi over 2x to including arc tangent of this thing. And this piece from the intersection to 0 from arc tangent of a plus square root of a square minus 4 over 2x, not including 0. I'm not sure which is plus and which is minus. Let's just think about it. If a is negative, then this is more negative, right? Then this. So that's correct, right? So it's two intervals here. Now, what if a is in between? Then the whole solution actually is not working because this square root cannot be extracted. But if a is in between, all you can say is that for minus 2 a plus 2, you have x from minus pi over 2 to 0. So all this part, if a is in between these two values, then it doesn't really intersect the graphs, which means everything below is this whole piece. So that's what we have as a solution. So these three cases. Now, how to get the complete solution? Add pi and to each one of those, basically, because you have to add the period. So this is a complete solution. Depending on a, we could come up with different formulas, different intervals, and then we added the period. But the principle is basically the same. So if I would like to find out where some expression is, let's say, less or equal to a, I try to graph this expression more or less approximately. I'm not saying it's exact. And then intersect it with y equals to a and see on which side values are located in this case. Okay. That's mobile. Well, as you see, it's not really very difficult, but it's involved. There are certain calculations and you have to consider certain cases. It requires certain accuracy and I'm not sure I'm perfect in this, but anyway, trying. Okay. Now, next one. Six cosine square x plus five sine x less than or equal to seven. Again, the problem is we have a mixture of different trigonometric functions. However, it's obvious that cosine square can be converted into sine square and then you will get a situation which is relatively similar to whatever was in the previous problem. We have a quadratic inequality, so it's six minus six sine square of x plus five sine of x less than or equal to seven. So it's a quadratic inequality which we know how to solve. We will call sine of x v and what I will have is it would be minus six v square plus five v minus one because this is seven, this is six less than or equal to zero. Or I like to have plus at the square, so I will have six v square minus five v plus one greater than or equal to zero. I reverse the sine of the left part and reverse the equation to the opposite one. Now, what is this? This is a parabola, right? Directed upwards. So all we need to know are two values where it intersects the x-axis. So it's basically the solutions to the equation, to this equation. Alright, so what are the solutions? Twelve, five plus minus square root twenty-five minus four times twenty-four. So this is five plus minus one over twelve. So I have, if it's minus it's four, so it's one-third. And one-half. One-half and one-third. Two solutions. How can I use this? Well, basically if I'm looking for something of this particular polynomial to be greater than zero, it's quadratic polynomial, it's parabola. Greater means it's left from the left solution and to the right from the right solution, right? Again, parabola. So it goes to the left and to the right of this. So that's where my parabola is positive. So what it goes to is the following. It's sin of x is less or equal than the smallest one. That's one group of solutions and sin of x is greater or equal than one-half. That's another group of solutions, right? Now let's go to the graphic. Now it's easier. So from minus pi to pi, sin goes like this, right? This is minus one. This is one. And we're interested in one-half and one-third. So one group of solutions is less than one-third. Now this is one-third. This is one-half. So less than one-third is everything from minus pi to this point and from this point to pi, right? So from minus pi to this point, this is obviously arc sin of one-third. That's one interval and the second interval greater than this particular point. Now this particular point is pi minus arc sin of one-third and ends at pi. By the way, I specify the interval as two numbers, the left and the right part, and the square bracket means it's including the endpoints. If it's not including, like in case of tangent, I use the round parenthesis. Now what about one-half? That's yet another solution. So here we have greater than. So it means it's in between these two things, from here to here. So this is obviously arc sin of one-half and this one is pi minus arc sin of one-half. So these three intervals within, so it's this one and then the middle one and the small one, these three intervals represent the solution on the interval of periodicity from minus pi to pi. So to get the complete solution, you have to add two pi and two each one of them. Both ends of this. Well, that's it. By the way, in this case, as in the previous case, what I did, I have reduced my original inequality which involved different trigonometric functions to a quadratic inequality in terms of one particular trigonometric function. In case the first one was tangent, the second one was sine, but in both cases I have obtained, from analyzing the quadratic inequality, I have obtained the condition, a very simple condition in this case, on the original trigonometric function, on tangent or sine. And that I have converted into the interval for x, which was supposed to be. Next one. It's kind of similar, by the way, so far. This was also be similar. Now, as usually I don't remember the formula for sine of three-x, but I do remember it's something like some number multiplied by sine x and some number multiplied by sine of three-x. So I'm going to obtain this formula right now, but why am I talking about this? Well, because if this is a combination of sine and sine cubed of x, multiplied by sine, it will be sine square and sine to the fourth degree, which brings me to the point that it will be almost a quadratic equation, but instead of having just an independent variable, I will have sine square as an intermediate variable. So let's see, first of all, I don't remember, as I said, sine of three-x. Let me just derive it. So it's sine of x plus 2x, right? So it's sine of x times cosine of 2x, which is cosine square x minus sine square x. Plus cosine of x times sine of 2x, which is 2 sine x cosine x. And cosine square I will also convert into 1 minus sine square. 1 minus sine square x. So I have here 1 minus sine, minus 2 sine square. So if I multiply by sine, it will be sine x minus 2 sine cubed x. Plus 2 sine x times cosine square, which is 1 minus sine square equals sine square, sine x and 2 sine x will be 3 sine x minus 2 sine cubed and minus 2 sine cubed. So it's minus 4 sine cubed. That's where it is. Okay, so I will use it here. I will use this. This is completely invariant transformation. So I will have sine x times this, which is this, which would be 3 sine square x minus 4 sine to the 4x less than minus 760 by quadratic equation, right? So what I'm doing here right now is quite obvious. I will put sine square equals to v. And what do I have? Minus 4v square plus 3v. Let's put it here. Plus 716 less than or equal to 0, right? That's my equation. Well, again, I prefer to have the positive coefficient at v square. So I will have plus 4v square minus 3v minus 716. Greater than or equal to 0. So I reverse the sine here and I reverse the sine of inequality. Okay, first what we have to do, how to determine the points where this particular parabola is greater than 0, it's left to the left solution when it's equal to 0 and to the right from the right solution, right? We need two solutions of the corresponding equation. So it's 8, this is 3 plus minus square root of 9 minus, it's plus because it's minus, 4 times this times this, 4 times 4 is 16, this is 16, so it's just 7, right? 9 plus 7 is 16, square root is 4, so that's what I have, which is minus 1 eighths and 7 eighths. And this is sine square. Now, when this particular parabola is greater than 0, when v is less than or equal to minus 8 and v is greater than 7 eighths, right? Now, can sine square be less than 1 eighth? That's positive thing, or 0, but not less than minus 1 eighth, right? So this does not bring us any solutions, this one does. So we have to analyze basically the very simple inequality, sine square greater than 7 eighths. That's what we have reduced our problem to. Now, how to solve this? Well, there are many ways. Obviously, you can extract square root. Actually, I prefer to do something I think a little smarter. Remember, cosine of 2x is equal to cosine square minus sine square or 1 minus 2 sine square x, right? So sine square x is equal to what? 1 half, 1 minus cosine of 2x. Cosine square minus sine square. Cosine is 1 minus sine, so it's 1 minus 2 sine squares. And if I resolve it for sine square, it will be here. Cosine will be there. 1 minus, yes, so. And that's invariant transformation, so I'll do this. 1 half, 1 minus cosine of 2x greater than 7 eighths. Well, that's easier because there is no square here, right? OK, so it's 1 half minus 1 half cosine 2 x greater than 7 eighths times 8, right? So it's 8, no, that's 4. 4 minus 4 cosine of 2x greater than equal to 7, 4 cosine of 2x. It's here. It's less than or equal than minus 3. So cosine of 2x less than minus 3. And this we know how to solve for 2x, and then we'll just divide by 2, right? All right, so let's solve it, and that will be the end of it. Now, when is cosine of 2x less than minus 3 quarters? So let's say this is my cosine, 0 to 2 pi. This is pi, this is pi over 2, this is 3 pi over 2. Now minus 3 quarters is here. We need less than minus 3 quarters, right? So it's from this point to this point, this interval. Now, this is arc cosine of minus 3 quarters, right? Because arc cosine is defined on this. It's angle cosine of which is equal to minus 3 quarters. Now, I can actually do it this way. So it's from arc cosine of minus 3 quarters to this point. Now, if this is arc cosine of minus 3 quarters, this is 2 pi minus, because these are symmetrical, right? So 2 pi minus arc cosine of minus 3 quarters. So that's the solution, and adding 2 pi to both sides will give you the complete solution. Well, that's it. Was it difficult? I don't know, it's a little tedious, I would say, because you have to really first transform your original trigonometric inequality, which contains multiple trigonometric functions, into something more bearable with one trigonometric function, but not the simple one, but something like a quadratic inequality, where the one and only one trigonometric function is the argument. The first problem, there was a tangent. In the second problem, it was sine, and in the third problem, it was sine squared. And then, after you solve the quadratic inequality, basically by obtaining some kind of a condition on the argument, and argument is not original x, it's like tangent x or sine x or sine squared x, et cetera, you basically analyze the second wave of inequalities and the third if it's necessary, whatever. So by substitution, you reduce the problem further and further down to the elementary basic level of style like sine of x less than whatever. So that's the methodology. Hopefully it works. And all which is required for you is to see what kind of transformation you should do with the original inequality, which is a mixture of God knows what. How to transform it into something more palatable, more understandable, which can be resolved in steps. So you reduce your complex problem to two, three, four, whatever number, more simple problems. And that's basically the approach to an entire world. Whenever you have a complex problem, reduce it to a combination of small ones, and that would definitely do the trick. Well, that's why mathematicians are the best people who can solve the problems when nobody really knows what to do, because they're trained to think outside of the box. They don't know the algorithm, they come up with algorithms. And that's what this course is supposed to teach you, I hope. All right. Don't forget Unisor.com is the website which allows you to register, pass exams, enroll into different topics, basically to do this controlled educational environment. I definitely encourage you to register and to take exams, etc. So thank you very much. Try to repeat these calculations, these solutions yourself. That's very important. So you'll go again through the same problems. If you can do it by yourself, great. And good luck. Thank you.