$\begin{array}{l}
{\sin ^2}x + 2\sin x = 4\cos x + \sin 2x\\
\Leftrightarrow {\sin ^2}x + 2\sin x = 4\cos x + 2\sin x\cos x\\
\Leftrightarrow \sin x\left( {\sin x - 2\cos x} \right) + 2\left( {\sin x - 2\cos x} \right) = 0\\
\Leftrightarrow \left( {\sin x - 2\cos x} \right)\left( {\sin x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - 2\cos x = 0\\
\sin x = - 2(L)
\end{array} \right.\\
\Leftrightarrow \sqrt 5 \left( {\dfrac{1}{{\sqrt 5 }}\sin x - \dfrac{2}{{\sqrt 5 }}\cos x} \right) = 0\\
\Leftrightarrow \sqrt 5 \sin \left( {x - \alpha } \right) = 0\left( {\alpha = \arccos \dfrac{1}{{\sqrt 5 }}} \right)\\
\Leftrightarrow \sin \left( {x - \alpha } \right) = 0\\
\Leftrightarrow x - \alpha = k\pi \\
\Leftrightarrow x = \alpha + k\pi \left( {k \in \mathbb Z} \right)
\end{array}$
