2кос^2 х - кос х - 1=0

2cosx - cosx - 1 = 0

Пусть cosx = t, -1 t 1

Тогда: 2t - t - 1 = 0

D = b - 4ac = 1 - 4*2*(-1) = 1 + 8 = 9

t = (1 + 3)/2*2 = 4/4 = 1

t = (1 - 3)/2*2 = -2/4 = -1/2

1)cosx = 1

x = 2n, n Z

2) cosx = -1/2

x = +- arccos(-1/2) + 2n, n Z

x = +- ( - arccos1/2) + 2n, n Z

x = +- ( - /3) + 2n, n Z

x = +- 2/3 + 2n, n Z

Ответ: x = 2n, x = +- 2/3 + 2n, n Z.