Chapter 5, Problem 27RE

To determine

Solve the given trigonometric function

Answer to Problem 27RE

The general form of solution.

  $\begin{array}{l}x=\frac{\pi }{3}+2n\pi \hfill \\ x=\frac{2\pi }{3}+2n\pi \hfill \end{array}$

Where n is an integer.

Explanation of Solution

Given:

The trigonometric function,

  $\sin x=\sqrt{3}–\sin x$

By rewriting the trigonometric equation and isolating the trigonometric part on the left hand side.

  $\begin{array}{l} \sin x=\sqrt{3}–\sin x\hfill \\ \sin x+\sin x=\sqrt{3}\hfill \\ 2\sin x=\sqrt{3}\hfill \\ \sin x=\frac{\sqrt{3}}{2}\hfill \end{array}$

Because sin x has a period of 2p, so find the solution in $\left[0,2\pi \right).$

These solutions are $x=\frac{\pi }{3}andx=\frac{2\pi }{3}$

Finally add multiples of $2\pi$ to the obtained solutions in $\left[0,2\pi \right)$ to get the general form of solution.

  $\begin{array}{l}x=\frac{\pi }{3}+2n\pi \hfill \\ x=\frac{2\pi }{3}+2n\pi \hfill \end{array}$

Where n is an integer.