Giải phương trình:

$\sin 3x = \frac{{\sqrt 3 }}{2}$;

Do $\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}$ nên $\sin 3x = \frac{{\sqrt 3 }}{2}$$\Leftrightarrow \sin 3x = \sin \frac{\pi }{3}$

$\Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{3} + k2\pi \\3x = \pi - \frac{\pi }{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{9} + k\frac{{2\pi }}{3}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)$.