Giải thích các bước giải:
$\begin{array}{l}
\sin 3x + 2\cos 3x + \cos 2x - 2\sin 2x - 2\sin x - 1 = 0\\
\Leftrightarrow \left( {3\sin x - 4{{\sin }^3}x} \right) + \left( {4{{\cos }^3}x - 3\cos x} \right) + \left( {1 - 2{{\sin }^2}x} \right) - 4\cos x.\sin x - 2\sin x - 1 = 0\\
\Leftrightarrow \sin x - 2{\sin ^2}x - 4{\sin ^3}x + 4{\cos ^3}x - 3\cos x - 4\sin x.\cos x = 0\\
\Leftrightarrow \left( {\sin x + 2\cos x} \right)\left( {4\cos 2x - 2\sin x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + 2\cos x = 0\\
4\cos 2x - 2\sin x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{1}{{\sqrt 3 }}\sin x + \frac{2}{{\sqrt 3 }}\cos x = 0\\
4 - 8{\sin ^2}x - 2\sin x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x - \arccos \frac{2}{{\sqrt 3 }}} \right) = 0\\
\sin x = \frac{1}{4}\\
\sin x = \frac{{ - 1}}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \left( {\arccos \frac{2}{{\sqrt 3 }} + \frac{\pi }{2}} \right) + k2\pi \\
x = \left( {\pi - } \right)\arcsin \frac{1}{4} + k2\pi \\
x = \frac{{ - \pi }}{6} + k2\pi \\
x = \frac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\end{array}$
