Giải phương trình: sin5x + 2cos²x = 1.

sin5x + 2cos²x = 1

⇔ sin5x = 1 – 2cos²x

⇔ sin5x = –cos2x

⇔ sin5x = \(\sin \left( {\frac{\pi }{2} - \left( { - 2x} \right)} \right)\)

⇔ sin5x = \(\sin \left( {\frac{\pi }{2} + 2x} \right)\)

⇔ \(\left[ \begin{array}{l}5x = 2x + \frac{\pi }{2} + k2\pi \\5x = - 2x - \frac{\pi }{2} + k2\pi \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x = \frac{\pi }{6} + 2k\frac{\pi }{3}\\x = - \frac{\pi }{{14}} + k\frac{{2\pi }}{7}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).