Question

Найти сумму целых решений неравенства на отрезке [-2;2]
2sin^2(x) + 5cos(x) >= -1

Answer 1

2-2cos²x+5cosx+1≥0
2cos²x-5cosx-3≤0
cosx=a
2a²-5a-3≤0
D=25+24=49
a1=(5-7)/4=-1/2
a2=(5+7)/4=3
-1/2≤a≤3⇒-1/2≤cosx≤3
x∈[4π/3+2πn;8π/3+2πn]