Помогите решить уравнение 4*cos^2-3=0
4cos²х - 3 = 0 (2cosx - √3)(2cos x + √3) = 0 2cosx - √3 = 0 или 2cos x + √3 = 0 cosx = √3/2 cosx = -√3/2 x = +-arccos(√3/2) + 2πn, n ∈ Z x = +-arccos(-√3/2) + 2πm, m ∈ Z x=+-π/6 + 2πn, n ∈ Z x=+-5π/6 + 2πn, n ∈ Z