 Hi, I'm Zor. Welcome to a new Zor education. I'd like to start a new topic. It's related to trigonometric inequalities. Well, first of all, which inequality can be called trigonometric? Well, obviously, the ones which involve some trigonometric function. That's not a big deal. Now, let me just spend a few minutes talking about general approach to how to solve inequalities. So, let's say you have some function and you would like to solve inequality of this type. Now, we are talking about real function, real arguments, and real constant A. How this particular inequality in general can be approached to solve? Well, they will deal with smooth functions, continuous functions. So, if I will have the points which are solution to this particular equality, let's say points are x1, x2, etc., xn. So, there are n solutions to this particular equality, equation. Then, look at the graph. This is a function f of x, and this is A. And I found these points where function f of x is equal to A. x1, x2, x3. So, it's obvious that for a smoothly behaving function, we are considering only these smooth functions. If I have, let's say, these two points, then the function in between these points should be either less or greater than this particular level, right? So, if x2 is equal to A and x3 is equal to A, and there is no points in between where it's equal to A. So, these are two consecutive points where f of x is equal to A. Then in between it's either less or greater than zero. So, to find out where exactly these particular conditions are held, we have to find the points where the corresponding equation becomes an identity. And then examine all the intervals in between these points, less than x1 in this particular case, between x1 and x2, between x2 and x3, and greater than x3 in this particular example. So, this is an approach which we will take to solve trigonometric equations as well. So, if we have a trigonometric equation, we will try to solve the corresponding, if we have an inequality, we will solve the corresponding equation, and then we'll examine the intervals in between the solutions of this equation. It might be a little bit complicated by the fact that trigonometric functions are periodic, which means the same solutions to the same equations actually are numerous, countable number of times on each period, but that's not a big problem. We will just solve the equation on some interval, which is the basic interval, the basic interval periodicity, and then we'll just add the period to all the solutions which we will get within this particular interval. Okay, so that's some kind of introduction to this particular lecture. Now, let's go to particular examples, and so my purpose during this lecture to explain solutions of unicmologies like tangent effects less than equal to a cotangent sine and cosine, basically, four cases. That's number one. Number two, I will usually use less or equal, but it's trivially expendable to less or greater or greater or equal or any other things. The difference is whether I will or will not include the points exactly where the tangent or sine or cosine equals to a. All right, so I will start with tangent, and the reason is exactly the same as my starting with tangent for equation because interval of periodicity of tangent is the same as equal of monatonic behavior and existence of the inverse function, or tangent, right? So this is your tangent minus pi over 2, pi over 2, and the pi is the period, and this is exactly the interval where function tangent behaves monotonically. It's monotonically increasing. I will solve the equation, sorry, inequality on this period from minus pi over 2 to pi over 2, and then just add pi, which is pi, pi is a period, right? And in this case, it's very simple actually because wherever the A is positive or negative, let's say it's negative, for instance, this is A. So this point is dividing the whole interval of periodicity into two parts. On the left, I have below A, and on the right, from this point, this is the point. And to the right of this point, to the end of the period, I have greater than A. So solutions to this particular inequality are very simple. So this particular point, what is this point? This is solution to equation which is arc tangent A, right? So this is the point on the x-axis where tangent is equal to A. So what I can say about solutions to this particular inequality is that x should be, we need less than or equal. So it's greater than minus pi over 2. It's not equal to minus pi over 2 because it's undefined, and less than or equal to arc tangent of A. So this is an interval from this to this, where the function tangent is below or on the level of A. Now, the general solution obviously is like this. I add pi times n where n is an integer number to here and to here. And this is a countable number of intervals, this interval, then this interval, and this and this countable number of intervals, and each interval is a solution. So the union of all these intervals is the solution. Well, that's it. It's simple with tangent, right? Now, let's go to cot tangent. That's where it's as simple actually. The period of cot tangent is also pi, but it's better for me to examine the function cot tangent on an interval from 0 to pi. It's also an interval of periodicity. I mean, any interval which has the length of pi is an interval of periodicity. But why did I choose this particular one? Well, because cot tangent behaves monotonically. This is pi over 2. Now, on this particular interval from 0 to pi, cot tangent behaves monotonically. So my inequality, now this is, let's now use the positive a. So this point is arc cot tangent of a. One and only point where the equation has the solution. This is the solution. So on the left, from this point, my function cot tangent is above a. And on the right, it's below a. So if I need the solution to this particular inequality, I have to do it the following way. I have to beneath less than or equal. So from this, x less than pi. From this to this, not including the boundary. So the end point is not included here, but it is included in this case because there is an equal, equal is acceptable solution. All I have to do right now is, as before, add pi n to both sides. That's the general solution. It's exactly the same as in case of a tangent. It's a small interval within the period, and then I expand it to all the periods. And what's the kind of simplicity in case of tangent and cot tangent? Because the interval of periodicity, which in both cases is pi, is the same as an interval of monotonic behavior and existence of the inverse trigonometric function for any real value a. So no matter which a I choose, I can always find this particular solution to this particular equation. So there are no restrictions, basically, on a or anything else. And it's easy to deal with monotonic behavior. It's just one inequality. Okay, situation with sine and cosine is slightly more complex. And let me explain you why. Let's approach again this graphically. Now, in case of, well, let's say with sine. So I will consider sine on the minus pi to pi. Eh, not exactly symmetrical. Something like this. Minus pi to pi. Okay? This is sine. And we are looking for solutions to this particular inequality. Well, first of all, it's obvious that if a is one or above, then every x would be actually the solution because sine cannot be greater than one. So let me just do it. Case number one. If a is greater or equal to one, then x can be any real number. Now, second case. If a is less than minus one, if it's below this level, less than minus one, then no solutions because any x would be greater and be less cylindrical. And only if a is in between these two boundaries from minus one to one, we can talk about some meaningful solutions, non-trivial solutions. I mean, everything is meaningful, but non-trivial solutions. All right. So let's examine a couple of cases separately. And here is why. If a is less than zero, negative, the solution to this is one particular interval here, right? All of these are solutions. Now, if a is here, then I have actually two different areas within the period where my sine is less than a. It's this interval and this interval. The whole this period, the whole this interval, because from this to this, sine is below this line and above this. So let me do the third case and split it in two halves. If a is greater than zero, so a is above. I have two different intervals, right? Now, this point is obviously arc sine of a. So my solution is from minus pi, less than or equal to x to arc sine of a. That's one particular interval, right? And the second interval is, now this is exactly in length, this particular interval is equal to this one. So it's pi minus arc sine. So from pi minus arc sine of a x and 2 pi. These are two intervals we're talking about. And obviously I have to add 2 pi n because in case of a sine, the period is 2 pi, right? I have to add it here and here and here and here. So for a greater than zero, this is the solutions. Now, and for, yes, what I mean is a is still less or equal to 1. It's not just greater because these cases I have to. So it's from zero to a to 1. And this is from minus 1, you can put equals here, to zero. Now, what happens in this case? So it's this case when a is below zero, so I have only one interval. Now, what is this point? This point is arc sine, right? Because the interval of periodicity is from minus pi over 2 to pi over 2. This is interval of periodicity. And for a, this is arc sine. And it's negative, by the way. So how can I get to this point? Well, that's minus pi minus arc sine of a, right? I have to move to the right. Now arc sine is negative, so I have to subtract to move to the right. Two arc sine. So that's the solution. Minus pi minus arc sine, that gives me this point. And this is arc sine, this is this point. And again, I have to add 2 pi m in both cases. That seems to be it. Now let's go to cosine. Half of the cosine is like this. It's symmetrical. Also from minus pi to pi. This is pi over 2, and this is minus pi over 2. And the situation is basically very similar. It all depends on the value of a. If a is greater than 1, again, actually even equal as well, then x can be any real number. Second, if a is less than minus 1, no solutions. Now, the third case, if a is in between minus 1 and 1. So this is minus 1, and this is 1. So if a is somewhere here. But you see, because cosine is a symmetrical function, no matter where I put this line, a, here, and in this case I have this interval and this interval, where the function cosine is below a. Now if a is negative, I still have the same two intervals here. One interval where it's below a, and another. There's always two intervals. In the case of a sine, sometimes I have one interval, sometimes two. In this case, in case of a cosine, I always have two intervals. So what is the interval of periodicity I took? The minus pi to pi. What's the interval of monotonic behavior for a cosine? Well, we usually traditionally, to have arc cosine, we take it from 0 to pi. So that's where we should really look for a cosine, for arc cosine. This is arc cosine. Now in this case, this is arc cosine. So arc cosine is always from 0 to pi. Now, how can I identify the left point? It's always minus pi. So it's always minus pi. And the right point is minus arc cosine. If this is arc cosine, or this is arc cosine, then this is minus arc cosine. Minus arc cosine of a. So that's one interval. And the second interval is from arc cosine to pi. So these are two intervals which are describing all the values of x on this particular period from minus pi to pi, where the cosine is less than or equal to a. And again, as usually, I have to add 2 pi m. So that's the solution. So what we have done, we have, number one, analyzed just a general approach how to solve inequality using the corresponding equation. So instead of finding where it's less or greater or whatever, we will find where it's equal. And then all the points in between the equality are inequality. And we are definitely using the smoothness of the graph of the function. Now, then we have analyzed the behavior of functions, tangent, cotangent, sine, and cosine, and investigated how to find what exactly are the values where these functions are less than or equal to some value. And as I said before, to transfer it to another inequality like greater or greater or equal or strictly less, it's trivial. So I'm not going to spend time on this. What's interesting is, again, that tangent and cotangent are a little bit more convenient in terms of their interval of periodicity is exactly the same as interval of monotonic behavior. So there is only one point where the corresponding equation has a solution. So it's always from some edge of the interval of periodicity to this point or from this point to another edge of periodicity. And then you end the period. With the sine and cosine, it's slightly more difficult because there are one or two intervals within the period. And it's all related to the fact that interval of periodicity is not the same as interval of monotonic behavior for both sine and cosine. That's why they had to consider different cases. Well, that's it. I do recommend you actually to repeat again these little calculations, which I did. And what's most important is you have to solve the problems. And I will definitely have some problems following this particular lecture, which explains just general theory. There are numerous examples of how to apply this knowledge and how to train your mind to find specific ways to resolve this or that particular inequality. These are simple elementary ones. These are the end road. When you are given a particular inequality to solve, this is the very end. So whenever you will reduce your original inequality to this, then you basically understand how to do it. But to reduce it to this particular form, that would require some ingenuity. And that's why the problems are all about. And that's why actually the whole course is about to develop your creativity and ingenuity. Well, don't forget everything is on Unisor.com. It's a free site and you are encouraged to register because that would enable your supervisor or parent, maybe or yourself if you want to play this role, to enroll you in particular topic, which will further enable you to take exams. Exams are important because there is no explanation like I'm doing right now. What's the solution? It's just an exam and you can take it as many times as you want, but that would really kind of check the level of knowledge and expertise which you have developed. Thanks very much and good luck.