Đáp án:
$C. 4$
Giải thích các bước giải:
$\begin{array}{l}\sin3x - \sin x + \sin2x = 0\\ \Leftrightarrow 2\cos2x.\sin x + 2\sin x\cos x = 0\\ \Leftrightarrow \sin x(\cos2x + \cos x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\cos2x = - \cos x\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\cos2x = \cos(\pi - x)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\2x = \pi - x + k2\pi\\2x = x - \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{3} + k\dfrac{2\pi}{3}\\x =- \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{3} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\\ \end{array}$
