giải phương trình cos4x-sin2x=0

Đáp án:

$\left[\begin{array}{l}x= -\dfrac{\pi}{4} + k\pi\\x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\cos4x - \sin2x = 0$

$\Leftrightarrow 1 - 2\sin^22x - \sin2x = 0$

$\Leftrightarrow 2\sin^22x + \sin2x - 1 = 0$

$\Leftrightarrow \left(\sin2x + 1\right).\left(\sin2x - \dfrac{1}{2}\right) = 0$

$\Leftrightarrow \left[\begin{array}{l}\sin2x = -1\\\sin2x = \dfrac{1}{2}\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}2x= -\dfrac{\pi}{2} + k2\pi\\2x = \dfrac{\pi}{6} + k2\pi\\2x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x= -\dfrac{\pi}{4} + k\pi\\x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$