8cos^4x - 8 cos^2x - cosx+1=0

Помогите решить

Замена cos x = t

8t^4 - 8t^2 - t + 1 = 0

8t^2 * (t^2 - 1) - (t - 1) = 0

8t^2 * (t - 1) (t + 1) - (t - 1) = 0

(t - 1) (8t^2 (t + 1) - 1) = 0

t1 = cos x = 1; x1 = 2pi*k

8t^3 + 8t^2 - 1 = 0

8t^3 + 4t^2 + 4t^2 + 2t - 2t - 1 = 0

(2t + 1) (4t^2 + 2t - 1) = 0

t2 = cos x = - 1/2; x2 = + - 2pi/3 + 2pi*n

Решаем квадратное уравнение

4t^2 + 2t - 1 = 0

D/4 = 1 + 4*1 = 5

t3 = cos x = (-1 - √5) / 4; x3 = + - arccos ((-1 - √5) / 4) + 2pi*m

t4 = cos x = (-1 + √5) / 4; x4 = + - arccos ((-1 + √5) / 4) + 2pi*q