Lời giải:Ta có: \({\rm{cos}}2x + \sin x = \sqrt 3 \left( {\cos x - \sin 2x} \right)\)
\(\begin{array}{l}
\Leftrightarrow \cos 2x + \sin x = \sqrt 3 \cos x - \sqrt 3 \sin 2x\\
\Leftrightarrow \cos 2x + \sqrt 3 \sin 2x = \sqrt 3 \cos x - \sin x\\
\Leftrightarrow \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x = \frac{{\sqrt 3 }}{2}\cos x - \frac{1}{2}\sin x
\end{array}\)
\( \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) = \cos \left( {x + \dfrac{\pi }{6}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}2x - \dfrac{\pi }{3} = x + \dfrac{\pi }{6} + k2\pi \\2x - \dfrac{\pi }{3} = - x - \dfrac{\pi }{6} + k2\pi \end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\x = \dfrac{\pi }{{18}} + k\dfrac{{2\pi }}{3}\end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)
Chọn đáp án A.
