Solutions in an Interval

Transcript

0:00Introduction

Hello everyone! Let me start with a strange equation: sine x equals x over ten. Try to solve it with any formula you know. You cannot, right? Apart from x equal to zero, there is no neat way to write down even a single root. And yet, by the end of this video, you will know exactly how many roots it has, without solving it at all. Today we learn to find and count the roots of a trigonometric equation inside a given range. Such counting questions come up often in JEE, so let's begin.

0:34Solve first, then list the angles

So, how do we find the solutions that lie inside an interval? Step one: solve the equation for the simplest ratio, something like sine x equals k, where k is just a fixed number. Okay? Step two: list the angles that actually lie inside the given interval, one by one, not just the general solution. Right? Now let me ask you this. Picture the sine wave over one full turn, going up to one and down to minus one. Draw a flat line at height k, somewhere between minus one and one. How many times does the wave cut that line in one full turn? Two times, yes, exactly! Once on the way up, and once on the way down. So sine x equals k has exactly two solutions in each interval of length two pi, when mod k is less than one. The two vertical bars around k mean mod k, its size without the sign, and the little sign pointing left means less than. Two pi is one full turn, in radians. And look at the brackets: square at a, round at a plus two pi, so the left end is counted but the right end is left out. That matters, because sine x equals zero has three solutions in the closed interval zero to two pi: zero, pi and two pi. Clear? Now tan. Tan repeats after every pi, and in one period it climbs through every real value. So in an interval of length pi, how many times does tan x equal k? It takes the value k just once, that is correct! So tan x equals k has exactly one solution in each interval of length pi, again with one end left out, whatever the value of k. With me?

2:18When algebra cannot finish: draw both graphs

Now, what about our equation from the start? Sine x is on one side, and on the other side is some function g of x that is not trigonometric at all, like x over ten. Here step one fails, because we cannot reduce it to one simple ratio. Okay? So how can we count the roots without ever finding them? We draw them, yes! Sketch y equals sine x, and y equals g of x, on the same axes. At every root the two heights are equal, so the graphs meet there. Got it? So the number of solutions is the number of points where the graphs meet. We use this in example two. Right?

2:58Watch the edges of the interval

Now, two small things that cost marks. First, the end points. Square brackets, like zero to two pi here, mean a closed interval, so both ends are included. Okay? So tell me, how many solutions does cos x equals one have, in the closed interval zero to two pi? The answer is two, not just the one at zero. Cos x is one at x equal to zero, and again at x equal to two pi, and both ends are inside the closed interval. Many students write only zero, and lose the root on the edge. Clear? Second, some values are not allowed by the equation itself. For example, tan x is not defined at x equal to pi over two, because cos x is zero there. So if a term of the equation is undefined at some value, drop that value, even if your algebra gives it. With me?

3:50Example 1: 2 sin squared x - 3 sin x + 1 = 0 in 0 to 2pi

Let's do the first example. Find the number of solutions of two sine squared x, minus three sine x, plus one, equals zero, in the closed interval zero to two pi. The small two after sine means squared. Pause the video and try it first. Done? Okay, let's check. This is just a quadratic in sine x. It factorises as two sine x minus one, times sine x minus one, equals zero. Okay? So sine x equals one half, or sine x equals one. Both values lie in the range minus one to one, ends included, so both are possible. Right? First, sine x equals one half. One half is less than one in size, so we expect two solutions in one full turn: pi over six, and five pi over six. Both lie inside zero to two pi. Got it? So, two values of sine x, two solutions for each, which makes four, right? No, no, careful, that is the trap here! For sine x equals one, the value one is not less than one in size. The sine wave only touches its top once in each full turn, so in zero to two pi we get just pi over two. Clear? So the solutions are pi over six, pi over two, and five pi over six. The number of solutions is three. With me?

5:14Example 2: number of solutions of sin x = x/10 by graphs

Now the harder one, and it is our equation from the start. Find the number of solutions of sine x equals x over ten. Use the graphs, and pause the video to try it first. Done? Okay, let's check. Sine x never goes above one, or below minus one, so the modulus of sine x is at most one. Then the modulus of x over ten must also be at most one; the sign with a line under it means less than or equal to. So x lies from minus ten to ten. Beyond that, the line has left the band of the sine wave. Okay? Now, x equal to zero works, since sine zero is zero, and zero over ten is also zero. So that is our first root. Right? Now take positive x, where the line is above zero. Let me ask you this: for x between pi and two pi, can the two graphs meet? No, they cannot, because the sine is negative there, and the line is positive. So for positive x, the graphs can meet only where the sine is positive, on its arches. From zero to ten there are two such arches: zero to pi, and two pi to three pi. The round brackets mean the ends are left out. Got it? On the first arch, the sine starts out steeper than the line, so it rises above it. At pi it comes back to zero while the line is still above zero, so they cross once, which gives one more root. On the second arch, the sine starts below the line, climbs to one, above it, and falls back below it, so they cross two times. And this whole arch fits before ten, since three pi is about nine point four two, which is less than ten; the wavy equals sign means approximately. With me? So we have one root at zero, and three positive roots. That makes four roots in all, right? Oh no, careful, we forgot negative x! If x is a root, then minus x is also a root, because sine of minus x is minus sine x, and the line changes sign too. So the negative side is the mirror image, with three more roots. Clear? So the total is one plus three plus three, which is seven. See, we never solved the equation, and still we know it has exactly seven roots. That is the mystery from the start, solved. Okay?

7:42Quick revision: list angles, draw graphs, check end points

So, quick revision. First, solve for one ratio, then list the angles inside the interval. Second, for sine x equals g of x, sketch both graphs and count the meeting points. And third, check the end points, and drop values where a term is undefined. Now go and try the practice problems on this page. You will find them easy. All the best!