$2 \cos^2 x − 2 \cos x− 1 = 0$ Find the solutions if $0^\circ \le x < 360^\circ$ 
Find the solutions of $$2 \cos^2 x − 2 \cos x− 1 = 0$$  for all $0^\circ ≤ x < 360^\circ$. 

For $0^\circ \le x < 360^\circ$, I'm getting $x=111.5^\circ$ and $x=248.5^\circ$.
Is this correct? Thanks!

Let $y = \cos(x)$.  Then we have the following:
$$2y^2 - 2y - 1 = 0$$
Using the quadratic formula, we get the roots to be:
$$y = \cos(x) = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$$
And so:
$$x = \arccos(\frac{1}{2} \pm \frac{\sqrt{3}}{2}) $$
Since $\frac{1}{2} + \frac{\sqrt{3}}{2}>1$, we can ignore this solution.  Since both of your solutions satisfy $\cos(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}$, you are indeed correct.
