ПОМОГИТЕ СРОЧНО решите уравнение -√8 sin2x+2=0

[latex]- \sqrt{8} sin2x + 2 = 0 \\ sin2x = \frac{2}{\sqrt{8}} \\ sin2x = \sqrt{\frac{4}{8}} \\ sin2x = \frac{\sqrt{2}}{2}} \\ 2x = (-1)^k arcsin(\frac{\sqrt{2}}{2}) + \pi k \in Z\\ 2x = (-1)^k\frac{\pi}{4} + \pi k, k \in Z \\ x = (-1)^k\frac{\pi}{8} + \frac{\pi k}{2},k\in Z[/latex]