Giải phương trình sin4x = – cos2x

sin4x = – cos2x

⇔ sin4x + cos2x = 0

⇔ 2sin2xcos2x + cos2x = 0

⇔ cos2x(2sin2x + 1) = 0

⇔ \(\left[ \begin{array}{l}\cos 2x = 0\\\sin 2x = - \frac{1}{2}\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\\sin 2x = \sin \left( { - \frac{\pi }{6}} \right)\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\x = \frac{{ - \pi }}{{12}} + k\pi \\x = \frac{{7\pi }}{{12}} + k\pi \end{array} \right.\)