$sin2x=\dfrac{-\sqrt[]{2}}{2}=sin(-\dfrac{\pi}{4})$
$⇔$\(\left[ \begin{array}{l}2x=-\dfrac{\pi}{4}+k2\pi\\2x=\pi+\dfrac{\pi}{4}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{array} \right.\)$(k∈Z)$