Đáp án:
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} +k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin x - \sqrt3\cos x = 1\\ \Leftrightarrow \dfrac{1}{2}\sin x - \dfrac{\sqrt3}{2}\cos x = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(x - \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{3} = \dfrac{\pi}{6} +k2\pi\\x - \dfrac{\pi}{3} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} +k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
