Đáp án:
$\left[\begin{array}{l} x = \pm \dfrac{\pi}{3} + k2\pi\\ x = \pm \dfrac{\pi}{6}+ k2\pi\end{array}\right.(k\in \Bbb Z)$
Giải thích các bước giải:
$4\cos^2x - 2(\sqrt3 + 1)\cos x + \sqrt3 = 0$
$\Leftrightarrow (2\cos x - 1)(2\cos x - \sqrt3) = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = \dfrac{1}{2}\\\cos x = \dfrac{\sqrt3}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l} x = \pm \dfrac{\pi}{3} + k2\pi\\ x = \pm \dfrac{\pi}{6}+ k2\pi\end{array}\right.(k\in \Bbb Z)$
