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

Question

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

Answer 1

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<br>t = -1/2
cos(x/2) = -1/2
x/2 = +-2π/3 + 2πn, n ∈ Z
x = +-4π/3 + 4πn, n ∈ Z