Tìm góc lượng giác $x$ sao cho:

sin(x – 60°) = $- \frac{{\sqrt 3 }}{2}$;

Do $\sin \left( { - 60^\circ } \right) = - \frac{{\sqrt 3 }}{2}$ nên sin(x – 60°) = $- \frac{{\sqrt 3 }}{2}$⇔ sin(x – 60°) = sin(– 60°)

$\Leftrightarrow \left[ \begin{array}{l}x - 60^\circ = - 60^\circ + k360^\circ \\x - 60^\circ = 180^\circ - \left( { - 60^\circ } \right) + k360^\circ \end{array} \right.\left( {k \in \mathbb{Z}} \right)$

$\Leftrightarrow \left[ \begin{array}{l}x = k360^\circ \\x = 300^\circ + k360^\circ \end{array} \right.\left( {k \in \mathbb{Z}} \right)$.