Đáp án:
$\begin{array}{l}
\cos 2x + 2\sin x - 1 - 2\sin x.\cos 2x = 0\\
\Rightarrow \cos 2x - 2\sin x.\cos 2x + 2\sin x - 1 = 0\\
\Rightarrow \cos 2x\left( {1 - 2\sin x} \right) - \left( {1 - 2\sin x} \right) = 0\\
\Rightarrow \left( {1 - 2\sin x} \right)\left( {\cos 2x - 1} \right) = 0\\
\Rightarrow \sin x = \frac{1}{2}\\
\cos 2x = 1\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi \\
2x = \pi + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi \\
x = \frac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
