5cos^2x-2cosx-3=0 решите пожалуйста

5cos2X - 2cosX - 3 = 0 5*(2cos²x - 1) - 2cosx - 3 = 0 10cos²x - 2cosx - 8 = 0 делим на 2 5cos²x - cosx - 4 = 0 cosx = t 5t² - t - 4 = 0 D = 1 + 4*5*4 = 81 t₁ = (1 - 9)/10 t₁ = - 4/5 t₂ = (1 + 9)/10 t₂ = 1 1)  cosx = - 4/5 x₁ = (+ -)arccos(-4/5) + 2πk, k ∈Z 2) cosx = 1 x₂ = 2πn, n∈Z