Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \pm \dfrac{1}{2}\arccos\dfrac{6}{7}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$3\cos4x - \sin^22x + \cos2x - 2 = 0$
$\Leftrightarrow 3(2\cos^22x - 1) - (1-\cos^22x) + \cos2x - 2 = 0$
$\Leftrightarrow 7\cos^22x + \cos2x - 6 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos2x = -1\\\cos2x = \dfrac{6}{7}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x = \pi + k2\pi\\2x = \pm \arccos\dfrac{6}{7}+ k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \pm \dfrac{1}{2}\arccos\dfrac{6}{7}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
