Đáp án:
$1)\left[\begin{array}{l} x=\dfrac{\pi}{2}+ k 2 \pi(k \in \mathbb{Z})\\ x=\dfrac{7\pi}{6}+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ 2)x=\pm \dfrac{1}{3}\arccos \left(\dfrac{2}{3}\right)+\dfrac{\pi}{18}+ \dfrac{k 2 \pi}{3}(k \in \mathbb{Z}).$
Giải thích các bước giải:
$1)\sin \left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\\ \Leftrightarrow \sin \left(x-\dfrac{\pi}{3}\right)=\sin \left(\dfrac{\pi}{6}\right)\\ \Leftrightarrow \left[\begin{array}{l} x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+ k 2 \pi(k \in \mathbb{Z})\\ x-\Leftrightarrow \dfrac{\pi}{3}=\dfrac{5\pi}{6}+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{\pi}{2}+ k 2 \pi(k \in \mathbb{Z})\\ x=\dfrac{7\pi}{6}+ k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ 2)\cos \left(3x-\dfrac{\pi}{6}\right)=\dfrac{2}{3}\\ \Leftrightarrow 3x-\dfrac{\pi}{6}=\pm \arccos \left(\dfrac{2}{3}\right)+ k 2 \pi(k \in \mathbb{Z})\\ \Leftrightarrow 3x=\pm \arccos \left(\dfrac{2}{3}\right)+\dfrac{\pi}{6}+ k 2 \pi(k \in \mathbb{Z})\\ \Leftrightarrow x=\pm \dfrac{1}{3}\arccos \left(\dfrac{2}{3}\right)+\dfrac{\pi}{18}+ \dfrac{k 2 \pi}{3}(k \in \mathbb{Z}).$
