3cos2x=2cosx помогите !!

Решение3cos2x = 2cosx3*(2cos²x - 1) - 2cosx = 06cos²x - 2cosx - 3 = 0cosx = t6t² - 2t - 3 = 0D = 4 + 4*6*3 = 76t₁ = (2 - 2√19)/12t₁ = (1 - √19)/6t₂ = (1 +√19)/61) cosx = (1 - √19)/6x₁ =(1 - √19)/6 + 2πk, k∈Z2)  cosx = (1 + √19)/6x₂ = (1 +  √19)/6  + 2πn, n∈Z