Решите уравнение: 2 cos 2 х + cos х – 1 = 0. помогите ((

... = 2(2cos^2(x) - 1) + cos(x) - 1 = 4cos^2(x) + cos(x) - 3 = 0
|cos(x)| = |y| <= 1
4y^2 + y - 3 = 0
D = 1+ 48 = 7^2
y1 = (-1 - 7) / 8 = -1
y2 = (-1 + 7) / 8 = 3/4
cos(x) = -1
x1 = p + 2pk
cos(x) = 3/4
x2 = +-arccos(3/4) + 2pk, k E Z