Đáp án:
$\begin{array}{l}
\sin 5x - \cos \left( {x + \dfrac{{5\pi }}{2}} \right) = \sin \left( {2x - \dfrac{\pi }{2}} \right)\\
\Rightarrow \sin 5x + \sin x = - \cos 2x\\
\Rightarrow 2.\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x - x}}{2} = - \cos 2x\\
\Rightarrow 2.sin3x.cos2x + cos2x = 0\\
\Rightarrow cos2x.\left( {2\sin 3x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
2\sin 3x + 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
\sin 3x = - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
3x = \dfrac{{ - \pi }}{6} + k2\pi \\
3x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{ - \pi }}{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{7\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.
\end{array}$
