Giải phương trình: sin3x – sinx = 0.

sin3x – sinx = 0

⇔ sin3x = sinx

⇔\(\left[ \begin{array}{l}3x = x + k2\pi \\3x = \pi - x + k2\pi \end{array} \right.\)

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