Solving trigonometry equation Please help me understand how to solve this for $0\leq x\leq360 $
I seem to have a problem with equations with powers.

$$3\sin^2 x-3\cos^2x+\cos x-1=0 $$

thinking that I would start by simplifying:

$$3 (\sin^2 x- \cos^2x)+\cos x - 1=0 $$

How I wish the equation in the bracket was in form $\sin^2 x+ \cos^2x$ which is equal to 1.
I also tried to substitute $\sin^2 x=1- \cos^2x$

$3\sin^2 x-3\cos^2 x+ \cos x - 1= 0$
$6\cos^2 x-\cos x - 2 = 0$
By the quadratic formula 
$cos x = \dfrac{1 \pm \sqrt{1-4(6)(-2)}}{12} =\dfrac{-1\pm 7}{12}$
So $cos x = \frac{-1}{2}$ and $cos x = \frac{2}{3}$
