Answer
The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{6}$, $\frac{\pi }{3}$, $\frac{7\pi }{6}$, and $\frac{4\pi }{3}$.
Work Step by Step
We know that the period of sine function is $2\pi $. So, in the interval there are two values at which the sine function is $\frac{\sqrt{3}}{2}$. One is $\frac{\pi }{3}$. And the sine is positive in quadrant II, thus the other value is:
$\begin{align}
& \pi -\frac{\pi }{3}=\frac{3\pi -\pi }{3} \\
& =\frac{2\pi }{3}
\end{align}$
Therefore, all the solutions to $\sin 2x=\frac{\sqrt{3}}{2}$ are given by:
$\begin{align}
& 2x=\frac{\pi }{3}+2n\pi \\
& x=\frac{\pi }{6}+n\pi
\end{align}$
Or,
$\begin{align}
& 2x=\frac{2\pi }{3}+2n\pi \\
& x=\frac{\pi }{3}+n\pi
\end{align}$
Where n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$ and $n=1$. And the equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{\pi }{6}+n\pi \\
& =\frac{\pi }{6}+0\times \pi \\
& =\frac{\pi }{6}+0 \\
& =\frac{\pi }{6}
\end{align}$
$\begin{align}
& x=\frac{\pi }{6}+n\pi \\
& =\frac{\pi }{6}+1\times \pi \\
& =\frac{\pi +6\pi }{6} \\
& =\frac{7\pi }{6}
\end{align}$
Or,
The second equation is also computed by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{\pi }{3}+n\pi \\
& =\frac{\pi }{3}+0\times \pi \\
& =\frac{\pi }{3}+0 \\
& =\frac{\pi }{3}
\end{align}$
$\begin{align}
& x=\frac{\pi }{3}+n\pi \\
& =\frac{\pi }{3}+1\times \pi \\
& =\frac{\pi +3\pi }{3} \\
& =\frac{4\pi }{3}
\end{align}$
