Đáp án:
$\begin{array}{l}
\cos \left( {2x + \frac{\pi }{3}} \right) + \sqrt 3 \sin x + \cos x + 3 = 0\\
\Rightarrow \frac{1}{2}.\cos 2\left( {x + \frac{\pi }{6}} \right) + \frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x + \frac{3}{2} = 0\\
\Rightarrow \frac{1}{2}\cos 2\left( {x + \frac{\pi }{6}} \right) + \sin \left( {x + \frac{\pi }{6}} \right) + \frac{3}{2} = 0\\
\Rightarrow \left( {1 - 2{{\sin }^2}\left( {x + \frac{\pi }{6}} \right)} \right) + 2\sin \left( {x + \frac{\pi }{6}} \right) + 3 = 0\\
\Rightarrow 2{\sin ^2}\left( {x + \frac{\pi }{6}} \right) - 2\sin \left( {x + \frac{\pi }{6}} \right) - 4 = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin \left( {x + \frac{\pi }{6}} \right) = 2\left( {ktm} \right)\\
\sin \left( {x + \frac{\pi }{6}} \right) = - 1
\end{array} \right.\\
\Rightarrow x + \frac{\pi }{6} = - \frac{\pi }{2} + k2\pi \\
\Rightarrow x = \frac{{2\pi }}{3} + k2\pi
\end{array}$
