Решите уравнение sin 2x - cos 2x = tg x .
sin 2x - cos 2x = tg x (2tg x)/(1 + tg^2 x) - (1 - tg^2 x)/(1 + tg^2 x) = tg x tg^3 x - tg^2 x - tg x + 1 = 0 (tg x - 1)(tg^2 x - 1) = 0 tg x = +- 1 x = (pi)/(4) + (kpi)/(2), k in Z
\( x = \frac{\pi}{4} + \frac{k\pi}{2},\ k \in \mathbb{Z} \)