5cos^2x-2cosx-3=0 решите пожалуйста

5cos2X - 2cosX - 3 = 05*(2cos²x - 1) - 2cosx - 3 = 010cos²x - 2cosx - 8 = 0 делим на 25cos²x - cosx - 4 = 0cosx = t5t² - t - 4 = 0D = 1 + 4*5*4 = 81t₁ = (1 - 9)/10t₁ = - 4/5t₂ = (1 + 9)/10t₂ = 11)  cosx = - 4/5x₁ = (+ -)arccos(-4/5) + 2πk, k ∈Z2) cosx = 1x₂ = 2πn, n∈Z