Giải phương trình: \(\sin 2x + 3\cos x - 2{\sin ^2}x - 3\sin x = 0\).

\(\sin 2x + 3\cos x - 2{\sin ^2}x - 3\sin x = 0\)

\( \Leftrightarrow 2\sin x\cos x - 2{\sin ^2}x + 3\left( {\cos x - \sin x} \right) = 0 \Leftrightarrow \left( {2\sin x + 3} \right)\left( {\cos x - \sin x} \right) = 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = - \frac{3}{2}\left( L \right)}\\{\cos x = \sin x}\end{array}} \right. \Leftrightarrow \cos x = \sin x\)

\( \Leftrightarrow \cos x = \cos \left( {\frac{\pi }{2} - x} \right) \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{2} - x + k2\pi }\\{x = x - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow 2x = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)\).