Find the solutions of the equation 3cos(2 theta) - 5cos(theta) + 2 = 0 in the interval 0 < theta < 2pi.

3cos^2(theta) - 3sin^2(theta) - 5cos(theta) + 2 = 0

3cos^2(theta) + 3cos^2(theta) - 3 - 5cos(theta) + 2 = 0

6cos^2(theta) - 5cos(theta) - 1 = 0

delta = 25 + 24 = 49

cos(theta) = (5 - 7)/12 = -1/6 or cos(theta) = 1

theta = 0 or arccos(-1/6) or 2pi - arccos(-1/6) or 2pi