Answer
$x=\pi +2 \pi n; x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=2 \pi n$
Work Step by Step
Here, we have $ 2 \sin x \cos x=\sin x$
It has to be noted that $\sin x (2 \cos x-1)=0$
This gives: $\sin x=0 \implies x=\pi, 2 \pi$
The general solution for $\cos (x/2)=\cos (0)$ is $ (x/2)=2 n \pi \pm (0)$
$ 2 \cos x-1=0 \implies \cos x=\dfrac{1}{2}$
This gives: $ x=\dfrac{\pi}{3}; \dfrac{ 5\pi}{3}$
Hence, we have $x=\pi +2 \pi n; x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=2 \pi n$
