Решение
1-sin2x=2cos^2(x/2)
1 - sin2x - 2*[(1 + cosx)/2] = 0
1 - sin2x - 1 - cosx = 0
2sinxcosx + cosx = 0
сosx(2sinx + 1) = 0
1) cosx = 0
x₁ = π/2 + πk. k ∈ Z
2) 2sinx + 1 = 0
sinx = - 1/2
x₂ = (-1)^(n +1) * arcsin(1/2) + πn, n ∈ Z
x₂ = (-1)^(n +1) * (π/6) + πn, n ∈ Z
Ответ: x₁ = π/2 + πk. k ∈ Z, k ∈ Z ; x₂ = (-1)^(n +1) * (π/6) + πn, n ∈ Z