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