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

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.

Another way.

Rewrite our equation in the following form: $$\left(\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}\right)^x+1=\left(\frac{2}{\sqrt{2+\sqrt{2}}}\right)^x.$$ We see the the left side decreases and the right side increases,

which says that our equation has one real root maximum.

But $2$ is a root and we are done!