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).
                                                 __
                                               √ 2
sin 2x cos x + cos 2x sin x = --------
                                                  2

Accepted Answer

Hello, today we're going to be solving the values of the trigonometric equation on the interval from 0 to 2 pi. So what we are given is sin of three X multiplied by cosine of X minus cosine of three X multiplied by sin of X. And that is equal to the square root of 3/2. Now, we can simplify this equation with the help of some trigonometric identities. And we have a trigonometric identity for the difference identity of sine, the difference identity of sine states that if you have sign of A minus B that is equal to sign of a multiplied by cosine of B minus cosine of a multiplied by sign of B. Now, if we take a look at the left hand side of the equation, notice that the first sine value and the first cosine value have the same angle and the second cosine value and second sine value have the same angle. What this means is that we can allow our A value to be three X and we can allow our B value to be X. And if we apply this to the trigonometric, I get identity going from right to left, we can rewrite the left hand side of the equation as sign of three X minus X equal to the square root of 3/2 three X minus X will give us the value of two X. So we have sine of two X equal to the square root of 3/2. Now, what we wanna do is we want to solve for sine of two X. But before we do this, let's apply a substitution to the angle of sine will allow Y to equal to two X. And we can rewrite the equation as sign of Y is equal to the square root 3/2. In order to answer this equation, we need to ask ourselves what angles of why can we plug in the sign to get the value of the square root 3/2? In order to answer this question, we're going to need to use a unit circle. So let's draw a unit circle. Now, on the range from 0 to 2 pi there are two values or two angles that give us a sign value of the square root 3/2. The first angle is located in quadrant one at the angle pi over three. The terminal point at this angle is one half comma square root 3/2. And the second angle is located in the second quadrant at the angle two pi over three. The terminal point at this angle is given to us as negative 1/ comma square over 3/2. So what this means is that the values of Y that make the equation true true are going to be Y is equal to pi over three and Y is equal to two pi over three. Now keep in mind that we allowed Y to equal to two X. So if we were to res substitute X back in for Y, we can rewrite both of the equations as two, X is equal to pi over three and two, X is equal to two pi over three. What we'll need to do is we'll need to solve for all the values of X that make this equation true. Now, since we are dealing with sine, we're going to have to add the period to pi to the values of the. And if we were to divide both of the equations by two, in order to solve for X, what we are left with is X is equal to pi over six plus pi and X is equal to pi over three plus pi. What this means is that the values of X or the initial values of X that makes sine of two X equal to the square of 3/2 true. Those initial values are going to be pi over six and pi over three. So the first two solutions for X are going to be pi over six and pi over three. And what these statements are saying is that we, if we want to find the remaining values of X, that makes sine of two X equal to the square or 3/2 true, we'll have to add the values of pi to pi over six and we'll have to add the value of pi to pi over three. If we add these values, we get X is equal to seven pi over six and X is equal to four pi over three. What this means is that the remaining values of X are going to be seven pi over six and four pi over three. So what this means is that we have four solutions of X that make the original equation true. Those solutions are going to be pi over six pi over 37 pi over six and four pi over three. 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). __ √ 2 sin 2x cos x + cos 2x sin x = -------- 2

Key Concepts

Trigonometric Identities

Double Angle Formulas

Interval Notation

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