Решите уравнение ((1 - tg^2 x)(1 + sin 2x))/((1 + tg^2 x)(1 - sin 2x)) = 3 + 2sin 2x - 2sin^2 x .

((1 - tg^2 x)(1 + sin 2x))/((1 + tg^2 x)(1 - sin 2x)) = 3 + 2sin 2x - 2sin^2 x cos 2x * (1 + sin 2x)/(1 - sin 2x) = 2 + 2sin 2x + cos 2x cos 2x * (2sin 2x)/(1 - sin 2x) = 2(1 + sin 2x) (sin 2x cos 2x - cos^2 2x)/(1 - sin 2x) = 0 (cos 2x(sin 2x - cos 2x))/(1 - sin 2x) = 0 [arrayl cos 2x = 0, sin 2x != 1 tg 2x = 1 array. 2x = -(pi)/(2) + 2kpi, (pi)/(4) + kpi, k in Z x = -(pi)/(4) + kpi, (pi)/(8) + (kpi)/(2), k in Z

\( x = -\frac{\pi}{4} + k\pi,\ \frac{\pi}{8} + \frac{k\pi}{2},\ k \in \mathbb{Z} \)