$sin(2x+\dfrac{\pi}{3})=-\dfrac{1}{2}$
$⇔ sin(2x+\dfrac{\pi}{3})=sin(-\dfrac{\pi}{6})$
$⇔ \left[ \begin{array}{l}2x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{7\pi}{6}+k2\pi\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\end{array} \right.$ $(k∈Z)$