And from that we see
$\displaystyle \tan{15^\circ} = \tan{2x}$
What do you think $\displaystyle x$ must be?
In fact, since the tangent function is positive in the first and third quadrants, we actually have
$\displaystyle \tan{(\{15^\circ, 195^\circ\} + 360^\circ n)} = 2 - \sqrt{3}$, where $\displaystyle n$ is an integer representing the number of times around the unit circle.
So solve $\displaystyle \tan{(\{15^\circ, 195^\circ\} + 360^\circ n)} = \tan{2x}$.
No.
You have $\displaystyle \{15^\circ, 195^\circ\} + 360^\circ n = 2x$
So $\displaystyle x = \{7.5^\circ, 97.5^\circ\} + 180^\circ n$.
Also, don't try and evaluate all possibilities. Since you do not have a restricted domain, you would need an infinity of them.
Just write the answer in terms of $\displaystyle n$.