Solving Quadratic Trig Equations Algebraically

Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 < x < 360°. Use exact solutions whenever possible.

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

I have done the following:

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

6cos^2(x) + 3cos(x) - 2cos(x) - 1 = 0

3cos(x) [2cos(x) + 1] -1 [2cos(x) +1] = 0

[3cos(x) -1] [2cos(x) + 1] = 0

3cos(x) - 1 = 0

3cos(x) = 1

cos(x) = 1/3

Quadrants I and IV

Reference Angle: ???

At this point I think I've done something wrong because there is no reference angle for cos(x) = 1/3.

2cos(x) + 1 = 0

2cos(x) = -1

cos(x) = -1/2

Quadrants II and III

Reference Angle= 60 degrees

180 - 60 = 120 degrees

180 + 60 = 240 degrees

Therefore, so far, I have solved for the values: 120 degrees and 240 degrees