Помогите! (

5cos2X-2cosX-3=0

5cos2X - 2cosX - 3 = 0

5 * (2cos ²x - 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