3 sin x/5 =2 cos^2 (x/5+П) помогите решить пожалуйста...

cos(π + x/5) = -cos(x/5) (-cos(x/5))^2 = cos^2(x/5) cos^2(x/5) = 1 - sin^2(x/5) 3sin(x/5) = 2 - 2sin^2(x/5) 2sin^2(x/5) + 3sin(x/5) - 2 = 0 t = sin(x/5) 2t^2 + 3t - 2 = 0 t = (-3 +- √(9 + 16)) / 4 = (-3 +- 5) / 4 t = -2 не подходит, так как cos(x/5) >= -1 t = 2/4 = 1/2 cos(x/5) = 1/2 x/5 = +-π/3 + 2πn, n ∈ Z x = +- 5π/3 + 10πn, n ∈ Z