Đáp án:
$-\dfrac{2\pi}{3}$
Giải thích các bước giải:
$\begin{array}{l}\quad \sqrt3\sin x-\cos x = 0\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin x - \dfrac12\cos x =0\\ \Leftrightarrow \sin\left(x - \dfrac{\pi}{6}\right) =0\\ \Leftrightarrow x - \dfrac{\pi}{6} = k\pi\\ \Leftrightarrow x = \dfrac{\pi}{6} + k\pi\quad (k\in\Bbb Z)\\ \text{Ta có:}\\ \quad -\pi \leq x \leq \pi\\ \to -\pi \leq \dfrac{\pi}{6} + k\pi \leq \pi\\ \to -\dfrac76 \leq k \leq \dfrac56\\ Do\,\,k\in\Bbb Z\\ nên\,\,k \in \{-1;0\}\\ \to \left[\begin{array}{l}x =- \dfrac{5\pi}{6}\\x = \dfrac{\pi}{6} \end{array}\right.\\ \to S = -\dfrac{5\pi}{6} + \dfrac{\pi}{6} = -\dfrac{2\pi}{3} \end{array}$
