Question

Помогите решить тригонометрическое уравнение, спасибо!

Answer 1

Решение
(sinx + 1)/(1 - cos2x) = (sinx + 1)/(1 + cos(π/2 + x))
(sinx + 1)/(1 - cos2x) = (sinx + 1)/(1 + sinx)
(sinx + 1)/(1 - cos2x) = 1

sinx + 1 = 1 - cos2x
1 - cos2x ≠ 0, cos2x ≠ 1, 2x ≠ 2πk, k ∈Z; x ≠  πk, k ∈Z

sinx  + cos2x = 0
sinx  + 1 - 2sin²x = 0
2sin²x - sinx - 1 = 0
sinx = t
2t² - t - 1 = 0
D = 1 + 4*2*1 = 9
t₁ = (1 - 3)/4
t₁ = - 1/2
t₂ (1 + 3)/4
t₂ = 1
1)  sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n ∈ Z
x₁ = (-1)^n* arcsin(-1/2) + πn, n ∈ Z
x₁ = (-1)^(n+1)* arcsin(1/2) + πn, n ∈ Z
x₁ = (-1)^(n+1)* (π/6) + πn, n ∈ Z
2) sinx = 1
x₂ =  π/2 + 2πm, m ∈ Z