Answer
The solutions of the equation are $x=\frac{\pi }{3}+2n\pi $ or $x=\frac{2\pi }{3}+2n\pi $, where n is any integer.
Work Step by Step
The x in $\sin x=\frac{\sqrt{3}}{2}$. Because $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$, so, the solutions of $\sin x=\frac{\sqrt{3}}{2}$ in [0,2 $\pi $ ) are:
$\begin{align}
& x=\frac{\pi }{3} \\
& =\pi -\frac{\pi }{3}
\end{align}$
$\begin{align}
& x=\frac{3\pi }{3}-\frac{\pi }{3} \\
& =\frac{2\pi }{3}
\end{align}$
The period of the sine function is $2\pi $, and the solutions are given by:
$x=\frac{\pi }{3}+2n\pi $ or $x=\frac{2\pi }{3}+2n\pi $, where n is any integer.
