# sin (x + pi/3) + sin (x - pi/3) = 1 State all of the solutions on the interval (0,2). What is the...

## Question:

{eq}sin (x + \frac{\pi}{3}) + sin (x - \frac{\pi}{3}) = 1 {/eq}

State all of the solutions on the interval {eq}[0,2) {/eq}. What is the strategy in determining which identities or technique to use in solving the equation?

## Fundamental Trigonometric Equations

Let {eq}a {/eq} be a constant number in the interval {eq}[-1,1] {/eq}.

To solve the equation {eq}\sin x=a {/eq}, first, find a specific angle {eq}y {/eq} such that {eq}\sin y=a {/eq}. Then, all the solutions of {eq}\sin x=a {/eq} is given by

To solve the equation {eq}\cos x=a {/eq}, first, find a specific angle {eq}y {/eq} such that {eq}\cos y=a {/eq}. Then, all the solutions of {eq}\cos x=a {/eq} is given by

Using the famous sum identities

{eq}\displaystyle \sin(a+b)=\sin a\cos b+\cos a\sin b,\quad \sin(a-b)=\sin a\cos b-\cos a\sin b, {/eq}

we have

{eq}\displaystyle \begin{align*} \sin\left(x+\frac{\pi}{3}\right)+\sin\left(x-\frac{\pi}{3}\right) &=\left(\sin x\cos \frac{\pi}{3}+\cos x\sin \frac{\pi}{3}\right)+\left(\sin x\cos \frac{\pi}{3}-\cos x\sin \frac{\pi}{3}\right)\\ &=2\sin x\cos \frac{\pi}{3}\\ &=\sin x. \end{align*} {/eq}

Therefore, the equation in the statement of the problem is equivalent to

{eq}\displaystyle \sin x=1. {/eq}

This equation has one solution in the interval {eq}[0,2) {/eq} and that is {eq}\boxed{x=\frac{\pi}{2}}. {/eq}