Giải phương trình:

$\sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{\pi }{6}} \right)$;

$\sin \left( {3x - \frac{\pi }{4}} \right) = \sin \left( {x + \frac{\pi }{6}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}3x - \frac{\pi }{4} = x + \frac{\pi }{6} + k2\pi \\3x - \frac{\pi }{4} = \pi - \left( {x + \frac{\pi }{6}} \right) + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}3x - x = \frac{\pi }{6} + \frac{\pi }{4} + k2\pi \\3x + x = \pi - \frac{\pi }{6} + \frac{\pi }{4} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{{12}} + k2\pi \\4x = \frac{{13\pi }}{{12}} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{24}} + k\pi \\x = \frac{{13\pi }}{{48}} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)$.