Answer
(a)$ x= n\pi, \ \text{or}\ \ x=\frac{2\pi}{3} +2n\pi,\ \text{or}\ \frac{4\pi}{3}+2n\pi $
(b) $ \bigg\{0,\frac{2\pi}{3} ,\pi, \ \frac{4\pi}{3} \bigg\} $
Work Step by Step
(a) Given
$$ \sin x+ 2\sin x\cos x =0 $$
Since
\begin{align*}
\sin x+ 2\sin x\cos x &=0\\
\sin x(1+ 2\cos x)& =0
\end{align*}
Then
$$\sin x=0,\ \ \text{or} \ \ 1+ 2\cos x=0$$
Case 1 , $$\sin x=0,\ \ \Rightarrow \ \ x=0,\pi ,2\pi,..= n\pi $$
Case 1 , $$\cos =\frac{-1}{2},\ \ \Rightarrow \ \ x=\frac{2\pi}{3} +2n\pi,\ \ \text{or} \frac{4\pi}{3} +2n\pi$$
Hence the general solutions is
$$ x= n\pi, \ \text{or}\ \ x=\frac{2\pi}{3} +2n\pi,\ \text{or} \frac{4\pi}{3} +2n\pi$$
(b) To find solutions on the interval $ 0\leq x<2\pi $, put values of $n $
$$ \bigg\{0,\frac{2\pi}{3} ,\pi , \frac{4\pi}{3} \bigg\} $$
