Đáp án:
$\left[\begin{array}{l}x = \pi + k2\pi\\ x = \arccos\dfrac{1}{5} + k\pi\\x = -\arccos\dfrac{1}{5} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$2\cos^2x + 4\cos x - 3\sin^2x + 2 = 0$
$\Leftrightarrow 2\cos^2x + 4\cos x - 3(1-\cos^2x) + 2 = 0$
$\Leftrightarrow 5\cos^2x + 4\cos x - 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\\cos x = \dfrac{1}{5}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\ x = \arccos\dfrac{1}{5} + k\pi\\x = -\arccos\dfrac{1}{5} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
