$sin(x+\dfrac{\pi}{3})=-\dfrac{\sqrt[]{3}}{2}=sin(-\dfrac{\pi}{3})$
$⇔$\(\left[ \begin{array}{l}x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=\pi+\dfrac{\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{2\pi}{3}+k2\pi\\x=\pi+k2\pi\end{array} \right.\) $(k∈Z)$