Giải phương trình\(\sin 2x + 2{\sin ^2}x - 6\sin x - 2\cos x + 4 = 0\).

\(\sin 2x + 2{\sin ^2}x - 6\sin x - 2\cos x + 4 = 0\)

\( \Leftrightarrow \left( {2\sin x\cos x - 2\cos x} \right) + \left( {2{{\sin }^2}x - 6\sin x + 4} \right) = 0\)

\(\begin{array}{l} \Leftrightarrow 2\cos x\left( {\sin x - 1} \right) + 2\left( {\sin x - 2} \right)\left( {\sin x - 1} \right) = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x + \cos x - 2 = 0} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = 1}\\{\sin x + \cos x = 2}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{2} + k2\pi }\\{\sin \left( {x + \frac{\pi }{4}} \right) = \sqrt 2 \left( {VN} \right)}\end{array}\left( {k \in \mathbb{Z}} \right)} \right.\\ \Leftrightarrow x = \frac{\pi }{2} + k2\pi ,k \in \mathbb{Z}\end{array}\)