Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
\cos 2x + \sqrt 3 .\sin 2x + \sqrt 3 .\sin x - \cos x + 4 = 0\\
\Leftrightarrow \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x + \frac{{\sqrt 3 }}{2}\sin x - \frac{1}{2}\cos x + 2 = 0\\
\Leftrightarrow \left( {\sin \frac{\pi }{6}.\cos 2x + \cos \frac{\pi }{6}.\sin 2x} \right) + \left( {\cos \frac{\pi }{6}.\sin x - \sin \frac{\pi }{6}.\cos x} \right) + 2 = 0\\
\Leftrightarrow \sin \left( {2x + \frac{\pi }{6}} \right) + \sin \left( {x - \frac{\pi }{6}} \right) + 2 = 0\\
\sin \left( {2x + \frac{\pi }{6}} \right) \ge - 1\\
\sin \left( {x - \frac{\pi }{6}} \right) \ge - 1\\
\Rightarrow \sin \left( {2x + \frac{\pi }{6}} \right) + \sin \left( {x - \frac{\pi }{6}} \right) + 2 \ge - 1 - 1 + 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\sin \left( {2x + \frac{\pi }{6}} \right) = - 1\\
\sin \left( {x - \frac{\pi }{6}} \right) = - 1
\end{array} \right.
\end{array}\]
