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).

𝝅 𝝅
sin ( x + ------ ) + sin ( x - ------ ) = 1
4 4

Accepted Answer

Hello, today we're gonna be solving the following trigonometric equation. So what we are given is sine of XX plus pi over three plus sine of X minus pi over three is equal to one. Now, we're going to need to simplify the given equation and we can do that by using the help of trigonometric identities. So we have a product to some identity which states that if we have the product of sine of a multiplied by cosine of B, this product is equal to one half multiplied by the quantity sine of A plus B plus sine of A minus B. Now, you are allowed to rewrite trigonometric identities. If we were to multiply both sides of this identity by two, we can rewrite the identity as two multiplied by S sign of a multiplied by cosine of B is equal to sine of A plus B plus sine of A minus B. Here. The right hand side of the identity matches the left-hand side of the given equation. So we can rewrite the left hand side of the given equation as two multiplied by sine of A multiplied by cosine of B. And this is where A is going to equal to X and B is going to equal to pi over three. So the left hand side of the given equation can be rewritten as two multiplied by sine of X multiplied by cosine pi over three is equal to one. Now, there's a bit of simplification that we can do here. First, let's go ahead and evaluate cosine pi over three and we can do this by using the help of a unit circle. The angle pi over three is located in the first quadrant and the terminal point of pi over three is given to us as one half comma the square root 3/2. Now recall that any terminal point on the unit circle is defined as cosine commas sign here cosine is any X value for any terminal point and sin is any Y value for any terminal point. So in order to evaluate cosine pi over three, we need to look at the X value for the terminal point of pi over three. And that value is given to us as one half. So cosine pi over three can be rewritten as one half. Now, if we were to multiply two and one half together, that is going to simplify to give us one and one multiplied by sin of X is going to leave us with sign of X. And the simplified form of the given equation is sign of X is equal to one. Now, we need to ask ourselves, what angles of X can we plug into sine within one rotation of the unit circle? That will give us the value of one. Well recall that sin is N Y value on the unit circle. And at the angle pi over two, the terminal point given to us at this angle is zero, comma one at this angle sign of X is equal to one. And this is the only angle that gives us a value of one within one rotation of the unit circle. So that means X is going to equal to pi over two. And with that being said, the answer to this problem is going to be D 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). 𝝅 𝝅 sin ( x + ------ ) + sin ( x - ------ ) = 1 4 4

Key Concepts

Trigonometric Identities

Sine Function Properties

Interval Notation

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