Verified answer

Решение

cos(π/2 - 2x) + sinx = 0

sin2x + sinx = 0

2sinx*cosx + sinx = 0

sinx * (2cosx + 1) = 0

1)  sinx = 0 

x = πk, k ∈ Z

2)  2cosx + 1 = 0

cosx = -1/2

x = ± arccos(-1/2) + 2πn, n ∈ Z

x = ±(π - arccos(1/2)) + 2πn, n ∈ Z

x = ±(π - π/3) + 2πn, n ∈ Z

x = ± (2π/3) + 2πn, n ∈ Z

Ответ: x = πk, k ∈ Z ; x = ± (2π/3) + 2πn, n ∈ Z.