Đáp án:
$S=\left\{-\dfrac{\pi}{4};-\dfrac{7\pi}{12}\right\}$
Giải thích các bước giải:
$\sin\left(2x+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}$
$⇒\left[ \begin{array}{l}2x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\,\,(k\in\mathbb Z)\\2x+\dfrac{\pi}{3}=-\dfrac{5\pi}{6}+k2\pi\,\,(k\in\mathbb Z)\end{array} \right.⇒\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\,\,(k\in\mathbb Z)\\x=-\dfrac{7\pi}{12}+k\pi\,\,(k\in\mathbb Z)\end{array} \right.$
Mà $x\in[-\pi;\pi]$
$⇒\left[ \begin{array}{l}x=-\dfrac{\pi}{4}\\x=-\dfrac{7\pi}{12}\end{array} \right.$
Vậy $S=\left\{-\dfrac{\pi}{4};-\dfrac{7\pi}{12}\right\}$.
