Question

5cos^2x-2cosx-3=0 решите пожалуйста

Answer 1

5cos2X - 2cosX - 3 = 0
5*(2cos²x - 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