2cos^2x-cosx-1=0 [0; 2п] 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 = 2π/3
ответ x = 2πn
x = +- 2π/3 + 2πn
2π; 2π/3;  4π/3