Question

tg x > sqrt 3

25 баллов.

Answer 1

$tgx\ \textgreater \ \sqrt3\; \; \Rightarrow \; \; \frac{\pi }{3}+\pi n\ \textless \ x\ \textless \ \frac{\pi }{2}+\pi n\; ,\; n\in Z$

Answer 2

tgx>√3
x=arctg√3
π/3+πk<x<π/2+πk;k€Z