Задача 45942 tg^3x+3>3tgx+tg^2x ...

marinichvika15

2020-04-03 21:00:10

tg^3x+3>3tgx+tg^2x

sova

2020-04-03 21:30:56

★

tg^3x+3>3tgx+tg^2x

tg^3x+3-3tgx-tg^2x>0
(tg^3x-tg^2x)+(+3-3tgx)>0
tg^2x(tgx-1)-3(tgx-1)>0
(tgx-1)(tg^2x-3)>0
(tgx-1)(tgx-sqrt(3))(tgx+sqrt(3))>0

___ (-sqrt(3)) __+__ (1) ___-__ (sqrt(3)) __+__

- sqrt(3) < tgx < 1 или tgx > sqrt(3)

[b]-(π/3)+πn <x<(π/4)+πn[/b] или [b] (π/3)+πn < x <(π/2)+πn, n ∈ Z[/b]