Đáp án:
$\left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x= k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$2(\cos x +\sqrt3\sin x)\cos x = \cos x - \sqrt3\sin x + 1$
$\Leftrightarrow 2\cos^2x - 1 + 2\sqrt3\sin x\cos x = \cos x - \sqrt3\sin x$
$\Leftrightarrow \cos2x + \sqrt3\sin2x = \cos x - \sqrt3\sin x$
$\Leftrightarrow \dfrac{1}{2}\cos2x + \dfrac{\sqrt3}{2}\sin2x = \dfrac{1}{2}\cos x -\dfrac{\sqrt3}{2}\sin x$
$\Leftrightarrow \cos\left(2x -\dfrac{\pi}{3}\right) = \cos\left(x +\dfrac{\pi}{3}\right)$
$\Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{3} = x + \dfrac{\pi}{3} + k2\pi\\2x - \dfrac{\pi}{3} =- x -\dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x= k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$
