Помогите решить cos(3x+\frac{\pi}{6} ) = - \frac{\sqrt{3}}{2}
Cos(3x+π/6)=-√3/2 3x+π/6=-5π/6+2πk U 3x+π/6=5π/6+2πk 3x=-π+2πk U 3x=2π/3+2πk x=-π/3+2πk/3 U x=2π/9+2πk/3,k∈z