Решить систему уравнений: Sin ( x / 2 ) - Cos 2y = 1 2 Sin ^2 (x / 2) - 3 Cos 2y = 2

Решите задачу:

$\sin ( x / 2 ) - \cos 2y = 1\\ 2 \sin ^2 (x / 2) - 3 \cos 2y = 2\\ \\ \sin (x/2) = 1 + \cos 2y\\ 2 (1 + \cos 2y)^2 - 3 \cos 2y = 2\\ 2 (1 + 2\cos 2y + \cos^2 2y) - 3 \cos 2y = 2\\ 2 + 4\cos 2y + 2\cos^2 2y - 3 \cos 2y = 2\\ \cos 2y + 2\cos^2 2y = 0\\ \cos 2y(1 + 2\cos 2y) = 0\\ \cos 2y_1 = 0 ; \cos 2y_2 = -1/2\\ \sin (x_1/2) = 1 ; \sin (x_2/2) = 1/2\\ \\ y_1 = \frac{\pi}{4}+\frac{\pi k}{2}\\ x_1 = \pi+4\pi k\\ \\ y_2 = \pi k \pm \frac{2\pi}{6}\\ x_2 = \pi + 4\pi k \pm \frac{2\pi}{3}$