Answer
Solutions of the equation are $\frac{\pi }{3}$ and $\frac{5\pi }{3}$.
Work Step by Step
On solving the equation,
$\begin{align}
& \sin x+2\sin \left( \frac{x}{2} \right)=\cos \left( \frac{x}{2} \right)+1 \\
& 2\sin \left( \frac{x}{2} \right)\cdot \cos \left( \frac{x}{2} \right)+2\sin \left( \frac{x}{2} \right)=\cos \left( \frac{x}{2} \right)+1 \\
& 2\sin \left( \frac{x}{2} \right)\left( \cos \left( \frac{x}{2} \right)+1 \right)-\left( \cos \left( \frac{x}{2} \right)+1 \right)=0 \\
& \left( \cos \left( \frac{x}{2} \right)+1 \right)\left( 2\sin \left( \frac{x}{2} \right)-1 \right)=0
\end{align}$
Simplified,
$\begin{align}
& \cos \left( \frac{x}{2} \right)=-1 \\
& \cos \left( \frac{x}{2} \right)=\cos \left( \frac{3\pi }{2} \right) \\
& x=3\pi
\end{align}$
But this does not lie in the interval $[0,2\pi )$. So, it will not be considered.
Further,
$\begin{align}
& \sin \left( \frac{x}{2} \right)=-\frac{1}{2} \\
& \sin \left( \frac{x}{2} \right)=\sin \left( \frac{\pi }{6} \right)
\end{align}$
Or
$\sin \left( \frac{x}{2} \right)=\sin \left( \frac{5\pi }{6} \right)$
This implies,
$x=\frac{\pi }{3},\frac{5\pi }{3}$
Thus, $\frac{\pi }{3}$ and $\frac{5\pi }{3}$ only lie in the interval $[0,2\pi )$.