Cos^2x+2=sin^2x+3cosx

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