How to solve this equation for $x$?$\left(\sqrt{2-\sqrt{3}}\right)^x + \left(\sqrt{2+\sqrt{3}}\right)^x = 2$

This is probably such a beginner question (and it's not homework). I've stumbled upon this:

$$\left(\sqrt{2-\sqrt{3}}\right)^x + \left(\sqrt{2+\sqrt{3}}\right)^x = 2$$

How to solve this equation for $x$? I've tried applying the $\log$ and also raising the expression to the power of 2, but couldn't go much further. I am probably missing something obvious.

The only real solution is given by $x=0$, since $\sqrt{2-\sqrt{3}}$ and $\sqrt{2+\sqrt{3}}$ are positive real numbers with product one, hence your equation is equivalent to: $$ e^{cx}+e^{-cx}=2 $$ or: $$ \cosh(cx) = 1 $$ for $c=\log\sqrt{2+\sqrt{3}}$.

HINT :

$$(2-\sqrt 3)^{\frac x2}+(2+\sqrt 3)^{\frac x2}-2=0\iff \left((2-\sqrt 3)^{\frac x4}-(2+\sqrt 3)^{\frac x4}\right)^2=0$$