(2cosx +√3) (2cos2x-3sinx-2) =0

\(\begin{array}{l}
pt \Leftrightarrow \left[ \begin{array}{l}
2\cos x + \sqrt 3 = 0\\
2\cos 2x – 3\sin x – 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = – \dfrac{{\sqrt 3 }}{2}\\
2\left( {1 – 2{{\sin }^2}x} \right) – 3\sin x – 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = – \dfrac{{\sqrt 3 }}{2}\\
4{\sin ^2}x + 3\sin x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = – \dfrac{{\sqrt 3 }}{2}\\
\sin x = 0\\
\sin x = – \dfrac{3}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{{3\pi }}{4} + k2\pi \\
x = k\pi \\
x = \arcsin \left( { – \dfrac{3}{4}} \right) + k2\pi \\
x = \pi – \arcsin \left( { – \dfrac{3}{4}} \right) + k2\pi
\end{array} \right.
\end{array}\)