Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{\cos x + \sqrt 3 \left( {\sin 2x + \sin x} \right) - 4\cos 2x.\cos x - 2{{\cos }^2}x + 2}}{{2\sin x - \sqrt 3 }} = 0\left( {DK:\sin x \ne \dfrac{{\sqrt 3 }}{2}} \right)\\
\Leftrightarrow \cos x + \sqrt 3 \left( {\sin 2x + \sin x} \right) - 4\cos 2x.\cos x - 2{\cos ^2}x + 2 = 0\\
\Leftrightarrow \cos x + \sqrt 3 \left( {2\sin x\cos x + \sin x} \right) - 4\left( {2{{\cos }^2} - 1} \right)\cos x - 2{\cos ^2}x + 2 = 0\\
\Leftrightarrow \sqrt 3 \sin x\left( {2\cos x + 1} \right) - 8{\cos ^3}x - 2{\cos ^2}x + 5\cos x + 2 = 0\\
\Leftrightarrow \sqrt 3 \sin x\left( {2\cos x + 1} \right) - \left( {8{{\cos }^3}x + 2{{\cos }^2}x - 5\cos x - 2} \right) = 0\\
\Leftrightarrow \sqrt 3 \sin x\left( {2\cos x + 1} \right) - \left( {2\cos x + 1} \right)\left( {4{{\cos }^2}x - \cos x - 2} \right) = 0\\
\Leftrightarrow \left( {2\cos x + 1} \right)\left( {\sqrt 3 \sin x - \left( {4{{\cos }^2}x - \cos x - 2} \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\cos x + 1 = 0\\
\sqrt 3 \sin x - \left( {4{{\cos }^2}x - \cos x - 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{ - 1}}{2}\\
\sqrt 3 \sin x + \cos x = 2\left( {2{{\cos }^2} - 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{ - 1}}{2}\\
\cos \left( {x - \dfrac{\pi }{3}} \right) = \cos 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi \\
x - \dfrac{\pi }{3} = 2x + k2\pi \\
x - \dfrac{\pi }{3} = - 2x + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \left( l \right)\\
x = - \dfrac{{2\pi }}{3} + k2\pi \left( c \right)\\
x = - \dfrac{\pi }{3} + k2\pi \left( c \right)\\
x = \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3}\left( c \right)
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3}
\end{array} \right.\,\left( {k \in Z} \right)
\end{array}$
