tg2x = - √3

Решим уравнение tg2x = - √3. $$tg2x = -\sqrt{3}$$ $$2x = arctg(-\sqrt{3}) + \pi n, n \in Z$$ $$2x = -\frac{\pi}{3} + \pi n, n \in Z$$ $$x = -\frac{\pi}{6} + \frac{\pi n}{2}, n \in Z$$ Ответ: $$x = -\frac{\pi}{6} + \frac{\pi n}{2}, n \in Z$$