Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{12} + k2\pi\\x = -\dfrac{5\pi}{12} + k2\pi\end{array}\right.\,\,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sqrt3\cos x - \sin x = \sqrt2\\ \Leftrightarrow \dfrac{\sqrt3}{2}\cos x - \dfrac{1}{2}\sin x = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin\dfrac{\pi}{3}\cos x - \cos\dfrac{\pi}{3}\sin x = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \sin\left(\dfrac{\pi}{3} - x\right)=\sin\dfrac{\pi}{4}\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{\pi}{3} - x = \dfrac{\pi}{4} + k2\pi\\\dfrac{\pi}{3} - x = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k2\pi\\x = -\dfrac{5\pi}{12} + k2\pi\end{array}\right.&(k \in \Bbb Z)\end{array}$
