Question

Решите неравенство: tgx≥√3

Answer 1

$tt tgxgeq sqrt{3}\arctgsqrt{3} +pi n leq xleq dfrac{pi}{2}+pi n,n in mathbb{Z}\\boxed{tt frac{pi}{3}+pi nleq xleq frac{pi}{2}+pi n,n in mathbb{Z}}$