2cos^2x-cosx-1=0 [0; 2 п] помогите решить и найти корни на отрезке

2cos^2x-cosx-1=0

cos x = t

2t^2 - t - 1 = 0

D = 1 + 8 = 9

t1 = (1 + 3) / 4 = 4/4 = 1

t2 = (1 - 3) / 4 = - 2/4 = - 1/2

cos x = 1

x = 2πn

cos x = - 1/2

x = + - 2π/3 + 2πn

при n = 1

x1 = 2π

x2 = - 2π/3 + 2π = 4π/3

при n = 0

x3 = 2 π/3

Ответ x = 2πn

x = + - 2π/3 + 2πn

2π; 2π/3; 4π/3