Помогите решить с1 ctgx+cos(pi/2+2x)=0

Решение
ctgx+cos(pi/2+2x)=0
ctgx-sin2x=0
cosx/sinx - 2sinxcosx = 0 * (sinx ≠ 0, x ≠ πk, k ∈ Z)
cosx - 2sin²xcosx = 0
cosx(1 - 2sin²x) = 0
1) cosx = 0
x = π/2 + πn, n ∈ Z
2) 1 - 2sin²x = 0
2sin²x = 1
sin²x = 1/2
sinx = - √2/2
x = (-1)^(n)(5π/4) + πn, n ∈ Z
sinx = √2/2
x = (-1)^(n)(π/4) + πn, n ∈ Z

Ответ: x = π/2 + πn, n ∈ Z; x = (-1)^(n)* (5π/4) + πn, n ∈ Z;
x = (-1)^(n)* (π/4) + πn, n ∈ Z