Question

sin²x-4cosx-1=0
Помогите пожалуйста

Answer 1

sin²(x) - 4cos(x) - 1 = 0

1 - cos²(x) - 4cos(x) - 1 = 0

-cos²(x) - 4cos(x) = 0

-cos(x) • (1 - 3cos(x)) = 0

• Система:

[ -cos(x) = 0

[ 1 - 3cos(x) = 0

[ cos(x) = 0

[ cos(x) = ⅓

[ x₁ = π/2 + 2πn, n ∈ ℤ

[ x₂ = ± arccos(⅓) + 2πn, n ∈ ℤ

Ответ:

x₁ = π/2 + 2πn, n ∈ ℤ

x₂ = ± arccos(⅓) + 2πn, n ∈ ℤ