Question

Срочно, пожалуйста!
1. 6cos^x +4 sin x cos x = 1
2.а) sin (x + п/3) ≤ √3/2; б) cos ( 3x - 2п/3) ≥ - 1/2

Answer 1

2. а) sin(x+π/3)≤√3/2
α₂=arcsin√3/2=π/3
α₁=-π-arcsin√3/2=-π-π/3=-4π/3
-4π/3+2πn≤x+π/3≤π/3+2πn | -π/3
-5π/3+2πn≤x≤2πn, n∈Z

б) cos(3x-2π/3)≥-1/2
α₂=arccos(-1/2)=2π/3
α₁=-arccos(-1/2)=-2π/3
-2π/3+2πn≤3x-2π/3≤2π/3+2πn | +2π/3
2πn≤3x≤4π/3+2πn | ×1/3
2πn/3≤x≤4π/9+2πn/3