Cos²x + 2 = Sin²x + 3Cosx
Cos²x + 2 = 1 - Cos²x + 3Cosx
Cos²x + Cos²x - 3Cosx + 2 - 1 = 0
2Cos²x - 3Cosx + 1 = 0
Cosx = t
2t² - 3t + 1 = 0
D = 3² - 4*2 = 9 - 8 = 1
t₁ = (3 + 1)/2*2 = 4/4 = 1
t₂ = (3 - 1)/2*2 = 2/4 = 1/2
Cox₁ = t₁ = 1 => x₁ = 2πn, n ∈ Z
Cosx₂ = t₂ = 1/2 => x₂ = ±π/3 + 2πn, n ∈ Z
Ответ: x₁ = 2πn, n ∈ Z, x₂ = ±π/3 + 2πn,n∈Z