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 - 2sin ²x = 0

2sin ²x = 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