Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{2} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin x -\sqrt3\cos x = 1$
$\to \dfrac12\sin x -\dfrac{\sqrt3}{2}\cos x =\dfrac12$
$\to \sin\left(x -\dfrac{\pi}{3}\right)=\sin\dfrac{\pi}{6}$
$\to \left[\begin{array}{l}x -\dfrac{\pi}{3}=\dfrac{\pi}{6} + k2\pi\\x -\dfrac{\pi}{3} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{2} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
