Помогите решить тригонометрическое уравнение, спасибо!

Решение(sinx + 1)/(1 - cos2x) = (sinx + 1)/(1 + cos(π/2 + x))(sinx + 1)/(1 - cos2x) = (sinx + 1)/(1 + sinx)(sinx + 1)/(1 - cos2x) = 1sinx + 1 = 1 - cos2x1 - cos2x ≠ 0, cos2x ≠ 1, 2x ≠ 2πk, k ∈Z; x ≠  πk, k ∈Zsinx  + cos2x = 0sinx  + 1 - 2sin²x = 02sin²x - sinx - 1 = 0sinx = t2t² - t - 1 = 0D = 1 + 4*2*1 = 9t₁ = (1 - 3)/4t₁ = - 1/2t₂ (1 + 3)/4t₂ = 11)  sinx = - 1/2x = (-1)^n*arcsin(-1/2) + πn, n ∈ Zx₁ = (-1)^n* arcsin(-1/2) + πn, n ∈ Zx₁ = (-1)^(n+1)* arcsin(1/2) + πn, n ∈ Zx₁ = (-1)^(n+1)* (π/6) + πn, n ∈ Z2) sinx = 1x₂ =  π/2 + 2πm, m ∈ Z