Question

решите уравнение tgx = ±√3

Answer 1

$\mathrm{tg}x = \pm\sqrt{3}$

$x = \mathrm{arctg}(\pm\sqrt{3})+\pi n$

$x = \pm\mathrm{arctg}\sqrt{3}+\pi n$

$\boxed{x = \pm\dfrac{\pi}{3}+\pi n,\ n\in\mathbb{Z}}$