6 cos2x + cosx-1=0

6 * cos^2 x + cos x - 1 = 0;

Пусть cos x = a, a ∈ [-1; 1].

6 * a^2 + a - 1 = 0;

Найдем дискриминант.

D = b^2 - 4 * a * c = 1^2 - 4 * 6 * (-1) = 1 + 4 * 6 = 1 + 24 = 25 = 5^2;

Вычислим корни.

a1 = (-1 + 5) / (2 * 6) = 4/12 = 1/3;

a2 = (-1 - 5) / 12 = - 6/12 = - 1/2;

Получаем:

1) cos x = 1/3;

x = + -arccos (1/3) + 2 * пи * n, n ∈ Z;

2) cos x = - 1/2;

x = + -arccos (-1/2) + 2 * пи * n, n ∈ Z;

х = + -2 * пи/3 + 2 * пи * n, n ∈ Z;

Ответ: x = + -arccos (1/3) + 2 * пи * n и х = + -2 * пи/3 + 2 * пи * n, n ∈ Z.