<s>[INST] $2 \cos^2 x − 2 \cos x− 1 = 0$ 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!
 [/INST] Observe that $2 \cos^2 x - 2 \cos x - 1$ is irreducible and cannot be factored.
Apply the quadratic formula:
$$\cos x = \frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\frac{2\pm\sqrt{12}}{4}=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$$
But $\frac{1 + \sqrt{3}}{2}$ is not in between $-1$ and $1$, so we can discard it accordingly. Thus,
$$\cos x = \frac{-1-\sqrt{3}}{2}$$
and so $$x=\cos^{-1} \bigg(\frac{-1-\sqrt{3}}{2}\bigg)\approx \boxed{111.5,248.5}$$
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