Answer
The solutions in the interval $[0,2\pi )$ are $0$, $\frac{\pi }{3}$, $\pi $ and $\frac{4\pi }{3}$.
Work Step by Step
The period of the sine function is $2\pi $. In the interval there are two values at which the sine function is $\frac{1}{2}$. One is $\frac{\pi }{6}$ . The sine is positive in quadrant II, thus the other value is:
$\begin{align}
& \pi -\frac{\pi }{6}=\frac{6\pi -\pi }{6} \\
& =\frac{5\pi }{6}
\end{align}$
All the solutions to $\sin \left( 2x+\frac{\pi }{6} \right)=\frac{1}{2}$ are given by:
$\begin{align}
& 2x+\frac{\pi }{6}=\frac{\pi }{6}+2n\pi \\
& 2x=2n\pi \\
& x=n\pi
\end{align}$
Or,
$\begin{align}
& 2x+\frac{\pi }{6}=\frac{5\pi }{6}+2n\pi \\
& 2x=\frac{4\pi }{6}+2n\pi \\
& x=\frac{2\pi }{6}+n\pi \\
& =\frac{\pi }{3}+n\pi
\end{align}$
Where n is any integer. The solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$, $n=1$. The equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=n\pi \\
& =0\times \pi \\
& =0
\end{align}$
$\begin{align}
& x=n\pi \\
& =1\times \pi \\
& =\pi
\end{align}$
The second equation is also calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{\pi }{3}+n\pi \\
& =\frac{\pi }{3}+0\times \pi \\
& =\frac{\pi }{3}+0 \\
& =\frac{\pi }{3}
\end{align}$
$\begin{align}
& x=\frac{\pi }{3}+n\pi \\
& =\frac{\pi }{3}+1\times \pi \\
& =\frac{\pi }{3}+\pi \\
& =\frac{\pi +3\pi }{3}
\end{align}$
$=\frac{4\pi }{3}$
