Đáp án:
$\begin{array}{l}
\sin 5x + \sin x = \cos 2x\\
\Rightarrow 2.\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x - x}}{2} - \cos 2x = 0\\
\Rightarrow 2.\sin 3x.cos2x - cos2x = 0\\
\Rightarrow \cos 2x.\left( {2\sin 3x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sin 3x = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.
\end{array}$
