Giải các phương trình sau: \(\cos \left( {2x - \frac{{3\pi }}{2}} \right) + \sqrt 3 \cos 2x + 1 = 0\) 

\(\begin{array}{l}
\cos \left( {2x - \frac{{3\pi }}{2}} \right) + \sqrt 3 \cos 2x + 1 = 0\\
 \Leftrightarrow \sin 2x + \sqrt 3 \cos 2x =  - 1\\
 \Leftrightarrow \frac{1}{2}\sin 2x + \frac{{\sqrt 3 }}{2}\cos 2x =  - \frac{1}{2}\\
 \Leftrightarrow \sin \left( {2x + \frac{\pi }{3}} \right) = \sin \left( { - \frac{\pi }{6}} \right)\\
 \Leftrightarrow \sin \left( {2x + \frac{\pi }{3}} \right) = \sin \left( { - \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{3} =  - \frac{\pi }{6} + k2\pi \\
2x + \frac{\pi }{3} = \frac{{7\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{\pi }{4} + k\pi \\
x = \frac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)