Exponential equation with double radical I'm trying ti solve this exponential equation:
$(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=2^x$.
Here my try:
$\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ and
$\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$
So i get this relation:
$\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}}$
Using the substitution $t=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ the equation can be written as 
$(t)^x+(\frac{1}{t})^x=2^x$.
And from this i wrote:
$t^{2x}-(2t)^x-1=0$.
Now i don't how to proceed. Any suggestions?
Thanks!

Hint:
$$\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}=\cos(45^\circ+30^\circ)=\cos75^\circ$$
Similarly, $$\dfrac{\sqrt{2+\sqrt3}}2=\sin75^\circ$$
Now for $0<A<90^\circ,(\cos A)^x,(\sin A)^x$ is deceasing fucntion 
