Đáp án:
$\left[\begin{array}{l} x=-\dfrac{k 2\pi}{3}(k \in \mathbb{Z})\\ x=\dfrac{\pi}{9}+\dfrac{k 2\pi}{9}(k \in \mathbb{Z})\end{array} \right.$
Giải thích các bước giải:
$\sin 3x-\sin 6x=0\\ \Leftrightarrow \sin 3x=\sin 6x\\ \Leftrightarrow \left[\begin{array}{l} 3x=6x+k 2\pi(k \in \mathbb{Z})\\ 3x=\pi-6x+k 2\pi(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} -3x=k 2\pi(k \in \mathbb{Z})\\ 9x=\pi+k 2\pi(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=-\dfrac{k 2\pi}{3}(k \in \mathbb{Z})\\ x=\dfrac{\pi}{9}+\dfrac{k 2\pi}{9}(k \in \mathbb{Z})\end{array} \right.$
