Решение
1-cos2x=sinx
1 - ( 1 - 2sin²x) - sinx = 0
2sin²x - sinx = 0
sinx(2sinx - 1) = 0
1) sinx = 0
x₁ = πk,k ∈Z
2) 2sinx - 1 = 0
sinx = 1/2
x = (-1)^n*arcsin(1/2) + πn, n∈Z
x₂ = (-1)^n*(π/6) + πn, n∈Z
Ответ: x₁ = πk,k ∈Z ; x₂ = (-1)^n*(π/6) + πn, n∈Z