Đáp án:
$\left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$3\sin^2x + 2\cos^4x - 2 = 0$
$\Leftrightarrow 3(1 - \cos^2x) + 2\cos^4x - 2 = 0$
$\Leftrightarrow 2\cos^4x - 3\cos^2x + 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos^2x = 1\\\cos^2x = \dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\cos x = 1\\\cos x = -1\\\cos x = \dfrac{\sqrt2}{2}\\\cos x = -\dfrac{\sqrt2}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x = \pi + k2\pi\\x = \pm\dfrac{\pi}{4} + k2\pi\\x = \pm \dfrac{3\pi}{4} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x =\dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{array}\right.\quad (k \in \Bbb Z)$
