trig equation $(\sqrt{\sqrt{2}+1})^{\sin(x)}+(\sqrt{\sqrt{2}-1})^{\sin(x)}=2$ Please help me to solve this trig equation.

$$(\sqrt{\sqrt{2}+1})^{\sin(x)}+(\sqrt{\sqrt{2}-1})^{\sin(x)}=2$$

 A: Two hints:


*

*$z+z^{-1}=2\Rightarrow z=1$.

*$\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)=1$.
Solution:

 Let us denote $z=\left({\sqrt{2}+1}\right)^{\frac12\sin x}$. 
 Since $\displaystyle \sqrt{2}-1=\frac{1}{\sqrt{2}+1}$, 
 our equation can be equivalently rewritten as  \begin{align}z+z^{-1}=2.\end{align} 
 This has the unique solution $z=1$, which in turn implies (since, fortunately, $\sqrt{2}+1\neq 1$) that $\sin x=0$. 
 Thus $x=\pi n$ with $n\in\mathbb{Z}$.

A: To solve
$$\left(\sqrt{\sqrt{2}+1}\right)^{\sin(x)}+\left(\sqrt{\sqrt{2}-1}\right)^{\sin(x)}=2,$$
let us rewrite as
$$\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)} + \left(\sqrt{2}-1\right)^{\frac{1}{2}\sin(x)}=2.$$
First consider the auxiliary equation
$$ y + \frac{1}{y} = 2.$$
You can verify this has only one solution, $y=1$.
Next, notice
$$ \frac{1}{\sqrt{2}+1} = \sqrt{2}-1.
$$
By laws of exponents, our original equation is then
$$\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)} + 
\frac{1}{\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)}}=2,
$$
which we recognize as the auxiliary equation with
 $y=\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)}$.
Therefore the solution to our original equation has 
$$
\left(\sqrt{2}+1\right)^{\frac{1}{2}\sin(x)}=1.
$$
Since $\sqrt{2}+1>1$, we must hve
$$ \frac{1}{2} \sin(x) =0.$$
The solutions are the integer multiples of $\pi$.
