Đáp án:
$\left[\begin{array}{l} x = \pm \arccos\left(\dfrac{-\sqrt3 - \sqrt{35}}{8}\right) + 2\pi\\x = \pm\arccos\left(\dfrac{-\sqrt3 + \sqrt{35}}{8}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}2\cos2x + \sqrt3\cos x = 0\\ \Leftrightarrow 2(2\cos^2x - 1) + \sqrt3\cos x = 0\\ \Leftrightarrow 4\cos^2x + \sqrt3\cos x - 2 =0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = \dfrac{-\sqrt3 - \sqrt{35}}{8}\\\cos x = \dfrac{-\sqrt3 + \sqrt{35}}{8}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l} x = \pm \arccos\left(\dfrac{-\sqrt3 - \sqrt{35}}{8}\right) + 2\pi\\x = \pm\arccos\left(\dfrac{-\sqrt3 + \sqrt{35}}{8}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
