(1+√2 sinx - cos2x)/(ctgx - 1) = 0
1+√2 sinx - cos2x = 0
ctgx - 1 ≠ 0, x ≠ π/4 + πk, k ∈ Z, sinx ≠ 0, x ≠ πn, n ∈Z
1) 1+√2 sinx - cos2x = 0
1+√2 sinx - (1 - 2sin²x) = 0
1+√2 sinx - 1 + 2sin²x = 0
√2 sinx + 2sin²x = 0
sinx*(√2 + 2sinx) = 0
√2 + 2sinx = 0
2sinx = - √2
sinx = - √2/2
x = (-1)^n * arcsin (- √2/2) + πm, m ∈ Z
x = (-1)^(n+1) * arcsin (√2/2) + πm, m ∈ Z
x = (-1)^(n+1) * ( π/4) + πm, m ∈ Z
Ответ: x = (-1)^(n+1) * ( π/4) + πm, m ∈ Z