<s>[INST] 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.

 [/INST] 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:
$$\begin{aligned}
 & \left(1-\cos ^2 \theta\right)(1+\cos \theta)^2=2-\cos ^2 \theta \\
 \Rightarrow \quad & \cos ^4 \theta+2 \cos ^3 \theta-\cos ^2 \theta-2 \cos \theta+1=0 \\
 \Rightarrow \quad & \left(\cos ^2 \theta+\frac{1}{\cos ^2 \theta}\right)+2\left(\cos \theta-\frac{1}{\cos \theta}\right)-1=0
\end{aligned}$$
$$\begin{aligned}
 & \cos \theta-\frac{1}{\cos \theta}=u \\
 \Rightarrow & u^2+2+2 u-1=0 \\
 \Rightarrow \quad & u=-1 \\
 \Rightarrow \quad & \cos ^2 \theta+\cos \theta-1=0 \\
 \Rightarrow \quad & \cos \theta=\frac{-1+\sqrt{5}}{2}
\end{aligned}$$
$$\Rightarrow \quad x=\sin \theta=\sqrt{\frac{\sqrt{5}-1}{2}}$$
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