rearrange

$3(\sin(2x) + 1) = -\sqrt{3} \cos(2x)$

note that $\sin(2x)+1 \ge 0 \implies \cos(2x) \le 0$

square

$9(\sin^2(2x)+2\sin(2x)+1) = 3\cos^2(2x) = 3[1 - \sin^2(2x)]$

$3(\sin^2(2x)+2\sin(2x)+1) = \cos^2(2x) = 1 - \sin^2(2x)$

$4\sin^2(2x) + 6\sin(2x)+2 = 0$

$2\sin^2(2x)+3\sin(2x)+1 = 0$

$[2\sin(2x)+1][\sin(2x)+1] = 0$

$\sin(2x) = -\dfrac{1}{2} \text{ and } \sin(2x) = -1$

note $-\pi < x < \pi \implies -2\pi < 2x < 2\pi$

take it from here?