Find all $x\in\mathbb{R}$ such that $\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$. Find all $x\in\mathbb{R}$ such that:
$$
\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x\,.
$$

Immediately we notice that $x=2$ satisfies the equation.
Then we see that $LHS=a^x+b^x$, where $a<1$ and $b<2$, therefore $RHS$ grows faster (for larger $x$, $LHS\approx b^x<2^x$)
Hence $x=2$ is the only real solution.

Unfortunately I don't know whether this line of reasoning is correct. Moreover, if it is indeed correct, how to write this formally?

It is not difficult (formula for the double angle) to show that $$\sin \left(\frac{ \pi }{8} \right)= \sqrt{ \frac{2- \sqrt{2} }{4} }$$ which in combination with the trigonometric one gives $$\cos^2\left(\frac{ \pi }{8} \right)=1-\sin^2\left(\frac{ \pi }{8} \right)=\frac{2+ \sqrt{2} }{4}\Rightarrow \cos \left(  \frac{ \pi }{8} \right)= \sqrt{  \frac{2+ \sqrt{2} }{4} }$$ thus our equation can be expressed equivalently in the form $$\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$$ $$\left( \sqrt{\frac{2- \sqrt{2} }{4}}\right)^x+\left( \sqrt{\frac{2+ \sqrt{2} }{4}}\right)^x=1 $$ $$\sin^x\left(  \frac{ \pi }{8} \right)+\cos^x \left(  \frac{ \pi }{8} \right)=1$$ of course thanks to the trigonometric one $x=2$ is a trivial solution. Uniqueness of this solution is due to the fact $\sin \& \cos \le 1$. Formally, you can consider cases $x>2$ or $x<2$ and estimate the left side.
