Giải phương trình:

\(\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{1}{2}\);

Do \(\sin \left( { - \frac{\pi }{6}} \right) = - \frac{1}{2}\) nên \(\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{1}{2}\)\( \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{6}} \right)\)

 \( \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{6} + k2\pi \\2x - \frac{\pi }{6} = \pi - \left( { - \frac{\pi }{6}} \right) + k2\pi \end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{4\pi }}{3} + k2\pi \end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\)

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