Exponential equation with double radical

Question

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!

Answer 1

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

Answer 2

Hint:

Write $s= \sqrt{2+\sqrt{3}}$, then $\sqrt{2-\sqrt{3}} = {1\over s}$

Then $$s^x+({1\over s})^x\geq 2 \implies x\geq 1$$

Answer 3

Hints:

1. Graph each of the equations above. Is there one that doesn't reach the origin?
2. Is there a point where the equations intersect? If so, is this the only place where it intersects?