Verified answer

Sin(3π/2 - x) = - cosx

Cos2x = 2cos²x - 1 . . . .. ........Вот она, вот она, фооормула..

2cos²x - 1 = - cosx

2cos²x + cosx - 1 = 0

cosx = t

2t² + t - 1 = 0

D = 1 + 8 = 9 = 3²

t₁ = (-1+3)÷4 = ½

t₂ = (-1-3)÷4 = -1

x = ±π/3 + 2πk, k ∈ Z.

x = ± π + 2πk, k ∈ Z.