Решение
Cos2x=1+sinx
1 - 2sin²x - sinx = 1
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
x = (-1)^(n + 1) *arcsin(1/2) + πn, n∈Z
x₂ = (-1)^(n + 1) *(π/6) + πn, n∈Z
Ответ: x₁ = πk, k∈Z ; x₂ = (-1)^(n + 1) *(π/6) + πn, n∈Z