Answer
$x=\frac{\pi}{4}+n\frac{\pi}{2}$ and $x=\frac{\pi}{2}+n\pi$, where $n$ is an integer.
Work Step by Step
$sin~2x~sin~x=cos~x$
$2~sin~x~cos~x~sin~x-cos~x=0$
$cos~x~(2~sin^2x-1)=0$
$sin~x=±\frac{\sqrt 2}{2}$ or $cos~x=0$
The solutions in $[0,2\pi)$ are:
$x=\frac{\pi}{4}$,
$x=\frac{3\pi}{4}=\frac{\pi}{2}+\frac{\pi}{4}$,
$x=\frac{5\pi}{4}=2\frac{\pi}{2}+\frac{\pi}{4}$,
$x=\frac{7\pi}{4}=3\frac{\pi}{2}+\frac{\pi}{4}$,
$x=\frac{\pi}{2}$ and
$x=\frac{3\pi}{2}=\pi+\frac{\pi}{2}$
General solution:
$x=\frac{\pi}{4}+n\frac{\pi}{2}$ and $x=\frac{\pi}{2}+n\pi$, where $n$ is an integer.
