$\begin{array}{l}
\left( {2\sin x + 1} \right)\left( {\cos 2x + 2\sin 2x - 10} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\sin x + 1 = 0\\
\cos 2x + 2\sin 2x - 10 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = - \dfrac{1}{2}\\
\cos 2x + 2\sin 2x = 10
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi \\
\cos 2x + 2\sin 2x = 10\left( 1 \right)
\end{array} \right.\\
\left( 1 \right):\cos 2x + 2\sin 2x\\
= \sqrt 5 \left( {\dfrac{1}{{\sqrt 5 }}\cos 2x + \dfrac{2}{{\sqrt 5 }}\sin 2x} \right)\\
= \sqrt 5 \sin \left( {2x + \alpha } \right)\left( {\alpha = \arccos \dfrac{2}{{\sqrt 5 }}} \right)\\
\Rightarrow \sqrt 5 \le \cos 2x + 2\sin 2x \le \sqrt 5 \\
\to \left( 1 \right)\,vô\,nghiệm\\
\Rightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
x \in \left( {0;4\pi } \right)\\
\Rightarrow \left[ \begin{array}{l}
0 < - \dfrac{\pi }{6} + k2\pi < 4\pi \\
0 < \dfrac{{7\pi }}{6} + k2\pi < 4\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0 < - \dfrac{1}{6} + 2k < 4\\
0 < \dfrac{7}{6} + 2k < 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{{12}} < k < \dfrac{{25}}{{12}}\\
- \dfrac{7}{{12}} < k < \dfrac{{17}}{{12}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
k = 1\\
k = 2\\
k = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{11\pi }}{6}\\
x = \dfrac{{7\pi }}{6}\\
x = \dfrac{{19\pi }}{6}\\
x = \dfrac{{23\pi }}{6}
\end{array} \right.\\
\to 4\, giá\, trị
\end{array}$
