Решить уравнение (подробно)
2sin² x/2=cos (3π/2+x/2)

2sin²(x/2) = cos(3π/2 + x/2)2sin²(x/2) = cos(π + π/2 + x/2)2sin²(x/2)  = sin(x/2) 2sin²(x/2) - sin(x/2) = 0sin(x/2)[2sin(x/2) - 1) = 01) sin(x/2) = 0x/2 = πn, n ∈ Zx = 2πn, n ∈ Z2) 2sin(x/2) - 1 = 0sin(x/2) = 1/2x/2 = (-1)ⁿπ/6 + πk, k ∈ Zx = (-1)ⁿπ/3 + 2πk, k ∈ ZОтвет: x = 2πn, n ∈ Z; (-1)ⁿπ/3 + 2πk, k ∈ Z.