Giải phương trình:

\(\sin 3x = \frac{{\sqrt 3 }}{2}\);

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

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

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