\(\begin{array}{l}
\quad \cos(3\sin x) = 0\\
\Leftrightarrow 3\sin x = \dfrac{\pi}{2} + k\pi\\
\Leftrightarrow \sin x = \dfrac{\pi}{6} + \dfrac{k\pi}{3}\\
\text{Ta lại có:}\\
\quad - 1 \leqslant \sin x \leqslant 1\\
\Leftrightarrow - 1 \leqslant \dfrac{\pi}{6} + \dfrac{k\pi}{3} \leqslant 1\\
\Leftrightarrow -\dfrac{6 + \pi}{\pi}\leqslant k \leqslant \dfrac{6-\pi}{2\pi}\\
\Rightarrow k = \{-1;0\}\\
+)\quad k = -1\ \text{ta được:}\\
\quad \sin x = - \dfrac{\pi}{6}\\
\Leftrightarrow \left[\begin{array}{l}x = - \arcsin\dfrac{\pi}{6} + n2\pi\\x = \pi + \arcsin\dfrac{\pi}{6} + n2\pi\end{array}\right.\\
+)\quad k = 0\ \text{ta được:}\\
\quad \sin x = \dfrac{\pi}{6}\\
\Leftrightarrow \left[\begin{array}{l}x = \arcsin\dfrac{\pi}{6} + n2\pi\\x = \pi - \arcsin\dfrac{\pi}{6} + n2\pi\end{array}\right.\\
\text{Vậy phương trình có các họ nghiệm là:}\\
x = \pm \arcsin\dfrac{\pi}{6} + n2\pi;\ x = \pi \pm \arcsin\dfrac{\pi}{6} +n2\pi\ \ \text{với}\ n\in\Bbb Z
\end{array}\)
