Solve for X:
3cos(x) + 3 = -2cos^2(x), for x E [-pi, 3pi]
$\displaystyle 2cos^2(x) + 3cos(x) + 3 = 0$
It might be easier for you to define y = cos(x), then look at:
$\displaystyle 2y^2 + 3y + 3 = 0$
You can use the quadratic equation to solve this, discard any solutions that are greater than 1 or less than -1, then take the inverse cosine.
The problem is that this quadratic has no real solutions. Thus you have no real solutions for x. Is there a typo perhaps?
-Dan