c) sin x = cos 3 x ⇔ [ x = π /2 − 3 x + k 2 π x = π − ( π /2 − 3 x ) + k 2 π , k ∈ Z .

c) \(\sin x = \cos 3x \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = \frac{\pi }{2} - 3x + k2\pi }\\{x = \pi - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi }\end{array},k \in \mathbb{Z}} \right..\)