Question

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

Answer 1

6 cos²x + cosx - 1 = 0
cosx = t
6t² + t - 1 = 0
D = 1² + 4 * 6 * 1 = 1 + 24 = 25
t1 = (-1 + 5)/2 * 6 = 4/12 = 1/3
t2 = (-1 - 5)/2 * 6 = -6/12 = -1/2
cosx = -1/2;
x = ±2/3π + 2πk;
cosx = 1/3;
x = arccos1/3 + 2πk;