$y=\sin2x+\sin\left(2x-\dfrac{2\pi}{3}\right)$
$=2\sin\left(2x-\dfrac{\pi}{3}\right).\cos\dfrac{\pi}{3}$
$=\sin\left(2x-\dfrac{\pi}{3}\right)\in [-1;1]$
Vậy:
$\min y=-1$ khi $\sin\left(2x-\dfrac{\pi}{3}\right)=-1\to x=\dfrac{-\pi}{12}+k\pi$
$\max y=1$ khi $\sin\left(2x-\dfrac{\pi}{3}\right)=1\to x=\dfrac{5\pi}{12}+k\pi$