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 $\sin^2 x=1- \cos^2x$

$3\sin^2{x} - 3\cos^2{x} + \cos{x} - 1 = 0$
$6\cos^2{x} - \cos{x} - 2 = 0$
$\cos{x} = -\frac{1}{2}\:\:\:\cos{x} = \frac{2}{3}$
