In Exercises 63–84, use an identity to solve each equation on ...

Question

In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅). cos 2x = cos x

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Hello, today we're going to be solving the following trigonometric equation on the interval from 0 to 2 pi. So what we are given is cosine of two X is equal to negative cosine of X. Since we are solving for the values of X, we want to go ahead and set the equation equal to zero. So we can do that by adding cosine of X to both sides of the equation. Now, one thing to clarify is that you cannot add cosine of two X and cosine of X together because the angles inside the cosine function are different. So if we were to add cosine of X to both sides of the equation, we can rewrite the equation as cosine two X plus cosine of X is equal to zero. Now, we'll need to simplify the equation and we'll need to simplify cosine of two X. Well, we have a trigonometric identity to help us do that. The trigonometric identity for cosine of two X states that cosine of two X is equal to two cosine squared of X minus one. So we can use this identity to rewrite cosine of two X and we can replace cosine of two X with two cosine squared of X minus one plus cosine of X is equal to zero. If we were to reorder the terms and put the cosine values first, we can rewrite the left hand side as two cosine squared of X plus cosine of X minus one is equal to zero. Now notice that the left hand side of the equation roughly resembles a polynomial. We can make this look like a polynomial. If we were to apply a substitution for cosine of X, let's go ahead and allow Y to equal to cosine of X. If we substitute cosine of X with Y, we can rewrite the left-hand side of the equation as two Y squared plus Y minus one is equal to zero. We can factor the left hand side to give us the quantities two Y minus one multiplied by Y plus one is equal to zero. Then we can set both of the factors of Y equal to zero using our zero property. So we have two separate equations. We have two Y minus one is equal to zero and we have Y plus one is equal to zero. On the left hand side, we're gonna go ahead and solve for Y. We can start by adding one to both sides of the equation leaving us with two Y is equal to one. And if we divide both sides of the equation by two, what we are left with is Y is equal to one half. And on the right hand side, in order to solve for Y, we're just going to subtract one of both sides of the equation leaving us with Y is equal to negative one. Now, if we res substitute cosine for Y, we get two separate equations for cosine, we have cosine of X is equal to a half and we have cosine of X is equal to negative one. What we need to ask ourselves in order to answer these questions is what angle of X can we plug in a cosine to give us the value of one half? And what angles of X can we plug in a sign to give us the values of negative one? While in order to answer this question, we need to use a unit circle. Now recall that any terminal point on the unit circle is defined as cosine comma S sign. This is where cosine is any X value of the terminal points and sin is any Y value of the terminal points. So let's start with the first equation cosine of X is equal to a half. What this means is what angles of X have one half in the X value of the terminal point while there's going to be two angles in total, the first angle lies in the first quadrant at pi over three. And the terminal point at this angle is one half comma square root 3/2. Since the terminal point has a half in its X value. That means that the solution for X is going to be pi over three. Now there's a second angle that has one half in the X value of the terminal point. And that angle is going to lie in the fourth quadrant of the unit circle. The angle is going to be five pi over three. And the terminal point at this angle is one half comma negative square root 3/2. Since the X value is a half at this angle, that means the second solution for X is going to be five pi over three. So those are going to be the two solutions of X for the first equation. Now we need to answer the second equation. And the second equation is asking us what angles have the terminal point or have the X value of negative one in the terminal point. Well, there's only one angle in the unit circle that has negative one in the X value of the terminal point. And that angle is going to be at pi the terminal point at pi is negative one comma zero. And since negative one is in the X value of the terminal point, that means that the solution of X to this equation is going to be pi. So to summarize, we have three solutions of X in total, we have X is equal to pi over three, we have X is equal to pi and we have X is equal to five pi over three. That is going to be the solutions of X in the range of 0 to 2 pi. And with that being said, the answer to this problem is going to be C so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅). cos 2x = cos x

Key Concepts

Double-Angle Identity for Cosine

Solving Trigonometric Equations

Interval Restriction and General Solutions

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