Đáp án:
$\begin{array}{l}
\left( {2\sin \,x + 1} \right)\left( {2cos2x - \sqrt 2 } \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin \,x = - \frac{1}{2}\\
cos2x = \frac{{\sqrt 2 }}{2}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{6} + k2\pi \\
x = \pi + \frac{\pi }{6} + k2\pi \\
2x = \frac{\pi }{4} + k2\pi \\
2x = - \frac{\pi }{4} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{6} + k2\pi \\
x = \frac{{7\pi }}{6} + k2\pi \\
x = \frac{\pi }{8} + k\pi \\
x = - \frac{\pi }{8} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
