Решение
1 - cosx - sin x/2 =0
1 - (1 - 2sin²x/2) - sinx/2 = 0
1 - 1 + 2sin²x/2 - sinx/2 = 0
2sin²x/2 - sinx/2 = 0
sinx/2(2sinx/2 - 1) = 0
1) sinx/2 = 0
x/2 = πk, k∈Z
x₁ = 2πk, k∈Z
2) 2sinx/2 - 1 = 0
sinx/2 = 1/2
x/2 = (-1)^narcsin(1/2) + πn, n∈Z
x/2 = (-1)^n(π/6) + πn, n∈Z
x₂ = (-1)^n(π/3) + πn, n∈Z
Ответ: x₁ = 2πk, k∈Z ; x₂ = (-1)^n(π/3) + πn, n∈Z