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

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