Помогите плиииз 8 cos^2 x/2+ 6 sin x/2-3=0

cos^2x = 1 - sin^2x 8 - 8sin^2(x/2) + 6sin(x/2) - 3 = 0 8sin^2(x/2) - 6sin(x/2) - 5 = 0 t = sin(x/2) 8t^2 - 6t - 5 = 0 t = (6 +- √(36 + 160)) / 16 = (6 +- 14) / 16 t = 20/16 не подходит, так как cos(x/2) <= 1 t = -1/2 cos(x/2) = -1/2 x/2 = +-2π/3 + 2πn, n ∈ Z x = +-4π/3 + 4πn, n ∈ Z