Đáp án:$\left[ \begin{array}{l}
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x = \frac{\pi }{6} + k\pi
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
\sqrt 3 \left( {\sin 3x - sinx} \right) = \cos 3x + \cos x\\
\Rightarrow \sqrt 3 .2.\cos \frac{{3x + x}}{2}.\sin \frac{{3x - x}}{2} = 2\cos \frac{{3x + x}}{2}.cos\frac{{3x - x}}{2}\\
\Rightarrow \sqrt 3 .\cos 2x.\sin x = \cos 2x.cosx\\
\Rightarrow cos2x.\left( {\sqrt 3 \sin x - \cos x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sqrt 3 \sin x - \cos x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{2} + k\pi \\
\sin \frac{\pi }{3}.\sin x - \cos \frac{\pi }{3}.\cos x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
\cos \left( {x + \frac{\pi }{3}} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x = \frac{\pi }{6} + k\pi
\end{array} \right.
\end{array}$
