$\begin{array}{l}
\sin 2x + \sin x - \frac{1}{{2\sin x}} - \frac{1}{{\sin 2x}} = 2\cot 2x\\
\Leftrightarrow {\rm{si}}{{\rm{n}}^2}2x + \sin x\sin 2x - \cos x - 1 = 2\cos 2x\\
\Leftrightarrow \left( {{{\sin }^2}2x - 1} \right) + 2{\sin ^2}x\cos x - \cos x - 2\cos 2x = 0\\
\Leftrightarrow - {\cos ^2}2x + \cos x\left( {2{{\sin }^2}x - 1} \right) - 2\cos 2x = 0\\
\Leftrightarrow - {\cos ^2}2x - \cos x\cos 2x - 2\cos 2x = 0\\
\Leftrightarrow \cos 2x\left( {\cos 2x + \cos x + 2} \right) = 0\\
\Leftrightarrow \cos 2x\left( {2{{\cos }^2}x + \cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
2{\cos ^2}x + \cos x + 1 = 0\left( {VN} \right)
\end{array} \right.\\
\Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2}
\end{array}$
