More rigorous method for this elementary problem? The problem is:

Find all real values of $x$ such that $$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x=2\sqrt{3}$$

One solution I received was as follows:

$5+2\sqrt{6}$ can be expressed as $(\sqrt{3}+\sqrt{2})^2$, and $5-2\sqrt{6}$ as $(\sqrt{3}-\sqrt{2})^2$. This means we want to solve $$(\sqrt{3}+\sqrt{2})^{2x}+(\sqrt{3}-\sqrt{2})^{2x}=2\sqrt{3}$$
  for $x$. Note that $$(\sqrt{3}+\sqrt{2})+(\sqrt{3}-\sqrt{2})=2\sqrt{3}$$
  so $x=\dfrac{1}{2}$ is a solution.

The problem with the solution is that it doesn't prove that $\dfrac{1}{2}$ is the only solution, not to mention its absolute inelegancy. Is there a more mathematically rigorous way of doing this?
I don't mind if higher algebra is used. I'm currently attempting to self-teach myself group theory, so including some of that in a solution would do me good. 
 A: Hint: $5+2\sqrt{6} = \dfrac{1}{5-2\sqrt{6}}$
A: Let $(5+2\sqrt6)^x=a$
As  $(5+2\sqrt6)(5-2\sqrt6)=1, (5-2\sqrt6)^x=a^{-1}$
$$\implies a+\frac1a=2\sqrt3\iff a^2-2\sqrt3a+1=0$$
$$\implies a=\frac{2\sqrt3\pm\sqrt{12-4}}2=\sqrt3\pm\sqrt2$$
If $a=\sqrt3+\sqrt2, (\sqrt3+\sqrt2)^{2x}=(\sqrt3+\sqrt2)^1$
If $a=\sqrt3-\sqrt2, (\sqrt3+\sqrt2)^{2x}=(\sqrt3-\sqrt2)^1=(\sqrt3+\sqrt2)^{-1}$ as $(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1$
What can we conclude if $a^x=a^y$ with $a\ne\pm1,0$
One observation: If $f(x)=a^x+\dfrac1{a^x}, f\left(-x\right)=f(x)$ 
So, if $\dfrac12$ is a solution, $-\dfrac12$ will also be a solution and of course vice versa. 
A: $\left(\sqrt{3}+\sqrt{2}\right)^{2x}+\left(\sqrt{3}-\sqrt{2}\right)^{2x}=\left(\left(\sqrt{3}+\sqrt{2}\right)^{x}+\left(\sqrt{3}-\sqrt{2}\right)^{x}\right)^2-2$
$\therefore \left(\sqrt{3}+\sqrt{2}\right)^{x}+\left(\sqrt{3}-\sqrt{2}\right)^{x}=\sqrt{2+2\sqrt{3}}\tag{1}$
Similarly,
$\left(\sqrt{3}+\sqrt{2}\right)^{x}-\left(\sqrt{3}-\sqrt{2}\right)^{x}=\sqrt{2\sqrt{3}-2}\tag{2}$
$\begin{align}\therefore (1)+(2) \implies \left(\sqrt{3}+\sqrt{2}\right)^{x} & =\dfrac{1}{2}\left(\sqrt{2\sqrt{3}+2}+\sqrt{2\sqrt{3}-2}\right)\\&= \dfrac{1}{\sqrt{2}}\left(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}\right)\end{align}$ 
$$\color{blue}{\boxed{\begin{align}\therefore \left(\sqrt{3}+\sqrt{2}\right)^{2x} & = \dfrac{1}{{2}}\left(2\sqrt{3}+2\sqrt{2}\right)=\left(\sqrt{3}+\sqrt{2}\right)\end{align}}}$$ 
A: Answer is attributed to ah-huh-moment's answer.
Since $5+2\sqrt{6}=\dfrac{1}{5-2\sqrt{6}}$ letting $y=5-2\sqrt{6}$ gives
$$\frac{1}{y^x}+y^x=2\sqrt{3}$$
which becomes
$$y^{2x}-2\sqrt{3}y^x+1=0$$
so
$$y^x=\sqrt{3}\pm \sqrt{2}$$
Plugging $y=5-2\sqrt{6}$ back in and taking $\log$ of both sides give
$$x=\log_{5-2\sqrt{6}}(\sqrt{3}\pm\sqrt{2})=\pm \frac{1}{2}$$
