Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:

Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$

My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^2-2\right)^2\\4x^4(1-x^2)=(x^4-x^2+1)^2$$
I don't know how can I proceed from here.
Is there a way so that not using the complicated expansion of polynomials?

I'm looking for methods that doesn't use $4$ or higher degree polynomial expansions.

 A: Firstly $x>0$
Let $\theta=\sin^{-1}{x} \Rightarrow x=\sin \theta$
We get
$$\sin \theta(1+\cos \theta)=\sqrt{1+\sin ^2 \theta} $$
Squaring both sides:
$$\begin{aligned}
 & \left(1-\cos ^2 \theta\right)(1+\cos \theta)^2=2-\cos ^2 \theta \\
 \Rightarrow \quad & \cos ^4 \theta+2 \cos ^3 \theta-\cos ^2 \theta-2 \cos \theta+1=0 \\
 \Rightarrow \quad & \left(\cos ^2 \theta+\frac{1}{\cos ^2 \theta}\right)+2\left(\cos \theta-\frac{1}{\cos \theta}\right)-1=0
\end{aligned}$$
$$\begin{aligned}
 & \cos \theta-\frac{1}{\cos \theta}=u \\
 \Rightarrow & u^2+2+2 u-1=0 \\
 \Rightarrow \quad & u=-1 \\
 \Rightarrow \quad & \cos ^2 \theta+\cos \theta-1=0 \\
 \Rightarrow \quad & \cos \theta=\frac{-1+\sqrt{5}}{2}
\end{aligned}$$
$$\Rightarrow \quad x=\sin \theta=\sqrt{\frac{\sqrt{5}-1}{2}}$$
A: We quickly observe that $x>0$.
Letting $1-x^2=u^2, \,u≥0$  we have:
$$x^2(1+2u+u^2)=1+x^2$$
This implies that,
$$\begin{align}&(1-u^2)(u^2+2u)=1\\
\implies &u(u+1)(u+2)(u-1)=-1\\
\implies &(u^2+u)(u^2+u-2)=-1\end{align}$$
Finally letting $t=u^2+u$, we have:
$$(t-1)^2=0\implies t=1$$
This leads to:
$$\begin{align}&u^2+u-1=0,\,u>0\\
\implies &u=1-u^2\\
\implies &x=\sqrt u=\sqrt {\frac{-1+\sqrt 5}{2}}~~~.\end{align}$$
A: Starting from: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ then
$$ x^2 \, (2 - x^2 + 2 \, \sqrt{1-x^2}) = 1 + x^2 \\
2 \, x^2 \, \sqrt{1-x^2} = 1 - x^2 + x^4 \\
4 \, x^4 \, (1-x^2) = 1 - 2 x^2 + 3 x^4 - 2 x^6 + x^8 \\
x^8 + 2 x^6 - x^4 - 2 x^2 + 1 = 0 \\
(x^4 + x^2 - 1)^2 = 0 \\
x^4 + x^2 - 1 = 0 \\ 
(x^2 + \alpha)(x^2 + \beta) = 0 \\
(x + i \sqrt{\alpha})(x - i \sqrt{\alpha})(x + i \sqrt{\beta})(x - i \sqrt{\beta}) = 0.$$
This gives the solutions as
$$ x \in \{ i \sqrt{\alpha}, - i \sqrt{\alpha}, i \sqrt{\beta}, - i \sqrt{\beta} \}, $$
or
$$ x \in \left\{ \frac{1}{\sqrt{\beta}}, - \frac{1}{\sqrt{\beta}}, \frac{1}{\sqrt{\alpha}}, - \frac{1}{\sqrt{\alpha}} \right\} $$
where $2 \, \alpha = 1 + \sqrt{5}$ (the golden ratio), and $2 \, \beta = 1 - \sqrt{5}$ (inverse golden ratio).
Now that there are values to consider it can be shown that $x \in \{ i \sqrt{\alpha}, - i \sqrt{\alpha} \}$ do not satisfy the original equation. Of the remaining values only one satisfies the equation, namely,
$$ x = \frac{1}{\sqrt{\alpha}} = i \, \sqrt{\beta} = \sqrt{\frac{\sqrt{5}-1}{2}} $$
