The side of a triangle are 2, 2, and $\sqrt{6} - \sqrt{2}.$  Enter the angles of the triangle in degrees, separated by commas.
By the Law of Cosines, the cosine of one of the angles is
\[\frac{2^2 + 2^2 - (\sqrt{6} - \sqrt{2})^2}{2 \cdot 2 \cdot 2} = \frac{4 \sqrt{3}}{8} = \frac{\sqrt{3}}{2},\]so this angle is $\boxed{30^\circ}.$  The other two angles must be equal, so they are $\boxed{75^\circ, 75^\circ}.$
By the Law of Cosines, the cosine of one of the angles is
\[\frac{2^2 + 2^2 - (\sqrt{6} - \sqrt{2})^2}{2 \cdot 2 \cdot 2} = \frac{4 \sqrt{3}}{8} = \frac{\sqrt{3}}{2},\]so this angle is $\boxed{30^\circ}.$  The other two angles must be equal, so they are $\boxed{75^\circ, 75^\circ}.$
The final answer is $$\boxed{30^\circ}$$. I hope it is correct.