Question

Теңсіздікті шешіңіз 5cosx+cos2x+3 ≥ 0

Answer 1

$5cosx+cos2x+3geq0\5cosx+2cos^2x-1+3geq0\2cos^2x+5cosx+2geq0\\cosx=tin[-1; 1]\\2t^2+5t+2geq0\Delta_t=5^2-4cdot2cdot2=25-16=9; sqrt{Delta_t}=sqrt9=3\\t_1=frac{-5-3}{2cdot2}=frac{-8}{4}=-2; t_2=frac{-5+3}{2cdot2}=frac{-2}{4}=-frac{1}{2}\\ [tin(-infty;-2] cup [-frac{1}{2}; infty)] cap [-1; 1]=[-frac{1}{2}; 1]$

$-frac{1}{2}leq cosxleq1\\cosxgeq-frac{1}{2} wedge cosxleq1\\O:xin[-frac{2pi}{3}+2kpi; frac{2pi}{3}+2kpi]; kinmathbb{Z}$