Answer
$\frac{\pi}{6}$, $\frac{\pi}{2}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$, $\frac{3\pi}{2}$, $\frac{11\pi}{6}$
Work Step by Step
$\tan \frac{1}{2}x+2\sin 2x=\csc x$
$\frac{1-\cos x}{\sin x}+2*2\sin x\cos x=\frac{1}{\sin x}$
Multiply both sides by $\sin x$:
$1-\cos x+4\sin^2x\cos x=1$
$-\cos x+4\sin^2x\cos x=0$
$\cos x(-1+4\sin^2x)=0$
$\cos x(4\sin^2x-1)=0$
$\cos x(2\sin x+1)(2\sin x-1)=0$
If $\cos x=0$, then the only solutions in $[0, 2\pi)$ are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
If $2\sin x+1=0$, then $2\sin x=-1$, ad $\sin x=-\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.
If $2\sin x-1=0$, then $2\sin x=1$, ad $\sin x=\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$.
