Тригонометрия. Задание в прикрепленном файле

$sin2x= \sqrt{3}cos2x \\ sin2x- \sqrt{3}cos2x =0 \ \ \ (\cdot \frac{1}{2} ) \\ cos \frac{\pi}{3}sin2x - sin \frac{\pi}{3}cos2x = 0 \\ sin(2x- \frac{\pi}{3})=0 \\ 2x= \pi k + \frac{\pi}{3} \\ x = \frac{\pi k}{2} + \frac{\pi}{6} \\ \boxed{k \in Z}$
Дан промежуток: $x \in [-\pi;0]$
$k = 0 \ ; \ x = \frac{\pi}{6} \approx \frac{3,14}{6} \notin [-\pi;0] \\ k = -1 \ ; \ x = -\frac{2\pi}{6} \approx -\frac{6,28}{6} \in [-\pi;0] \\ k=-2 \ ; \ x = -\frac{5 \pi}{6} \approx -\frac{15,7}{6} \in [-\pi;0] \\ k = -3 \ ; \ x = -\frac{8\pi}{6} \approx -\frac{21,98}{6} \notin [-\pi;0]$
Ответ: $-\frac{2\pi}{6} \ ; \ -\frac{5 \pi}{6}$