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

Question

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

Accepted Answer

Hello, today we're going to be solving the following trigonometric equation on the interval from 0 to 2 pi. So we are given sine of two X is equal to two cosine of X. In order to solve this equation, we'll first need to simplify sine of two X. In order to do this, we can use the following trigonometric identity sign of two X is equal to two sin of X cosine of X. So if we use this trigonometric identity, we can rewrite the left hand side of the equation as two sine of X cosine of X is equal to two cosine of X. Now, since we want to sell for X, we'll need to set the equation equal to zero. In order to do this, we can subtract two cosine of X to both sides of the equation. Doing so allows us to rewrite the equation as two sin of X cosine of X minus two cosine of X is equal to zero. Now notice that both of the terms on the left hand side have a multiple of two and a multiple of cosine, we can factor out two cosine of X from the left hand side and rewrite the equation as two cosine of X multiplied by the quantity sin of X minus one is equal to zero. We can then set both of the factors of X equal to zero. And we are left with two separate equations. Two cosine of X is equal to zero and sign of X minus one is equal to zero. On the left hand side, we can divide both sides of the equation by two. And that will leave us with cosine of X is equal to zero. And on the right hand side, we can add one to both sides of the equation, leaving us with sin of X is equal to one. So what we'll need to do is we'll need to use a unit circle to solve the following equations. Now recall that any terminal point found on the unit circle is defined as cosine comma sine. This is where cosine is any X value of the terminal point and sine is any Y value of the terminal points. So let's take a look at the first equation. We have cosine of X is equal to zero. And in order to answer this, we need to ask ourselves what angles of X can we plug in a cosine that will give us the value of zero? Well, in other words, what angles have zero in the exposition of its terminal point while there are two angles that have zero in the exposition of the terminal point. The first angle is at the angle pi over two. This angle has the terminal 0.0 comma one and the second angle will be three pi over two. And this angle has a terminal point of zero comma negative one. Now, both of these angles have zero in the terminal points or the exposition of the terminal points. So that means that there are two solutions to X in the first equation X is equal to pi over two and X is equal to three pi over two. Now let's go ahead and solve the second equation. The second equation states that sign of X is equal to one. So we need to ask ourselves what angles of X can we plug in a sign that will give us the value of one? Or in other words, what terminal points, what angles have one in the Y position of the terminal points? Well, there is only one angle that has positive one in the y position of its terminal point. And that's going to be the angle pi over two. So that means there is only one solution to the second equation X is equal to pi over two. So if you notice we have two repeating values for XX is equal to pi over two. And that means that there are only two solutions to X for the given equation X is equal to pi over two and X is equal to three pi over two and with that being said, the answer to this problem is going to be b 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).sin 2x = cos x

Key Concepts

Trigonometric Identities

Solving Trigonometric Equations

Unit Circle and Angle Measures

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