Đáp án:
$\left[\begin{array}{l}x=- \dfrac{\pi}{6}+ k2\pi\\x =\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}(\sin x - \cos x)^2 + \sqrt3\cos2x = 2\cos x + 1\\ \Leftrightarrow 1 - 2\sin x\cos x + \sqrt3\cos2x = 2\cos x + 1\\ \Leftrightarrow \sqrt3\cos2x - \sin2x = 2\cos x\\ \Leftrightarrow \dfrac{\sqrt3}{2}\cos2x - \dfrac{1}{2}\sin2x = \cos x\\ \Leftrightarrow \cos2x\cos\dfrac{\pi}{6} - \sin2x\sin\dfrac{\pi}{6} = \cos x\\ \Leftrightarrow \cos\left(2x + \dfrac{\pi}{6}\right) = \cos x\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{6} = x + k2\pi\\2x+\dfrac{\pi}{6} = - x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x=- \dfrac{\pi}{6}+ k2\pi\\x =\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
