3cos2x=2cosx

помогите!

Решение

3cos2x = 2cosx

3 * (2cos²x - 1) - 2cosx = 0

6cos²x - 2cosx - 3 = 0

cosx = t

6t² - 2t - 3 = 0

D = 4 + 4*6*3 = 76

t₁ = (2 - 2√19) / 12

t₁ = (1 - √19) / 6

t₂ = (1 + √19) / 6

1) cosx = (1 - √19) / 6

x₁ = (1 - √19) / 6 + 2πk, k∈Z

2) cosx = (1 + √19) / 6

x₂ = (1 + √19) / 6 + 2πn, n∈Z