<s>[INST] Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
 [/INST] Observe $\sqrt {1-x^2}≥0$ implies that,  $x≥0.$
Thus, under the restriction $0≤x≤\frac {\sqrt 2}{2}$ we have:
$$\begin{align}x\sqrt{1-x^2}&=\sqrt{1-x^2}-x\\
\left(x\sqrt {1-x^2}\right)^2&=1-2x\sqrt {1-x^2}\end{align}$$
Letting $x\sqrt {1-x^2}=u$, we have:
$$\begin{align}&u^2=1-2u\\
\implies &u^2+2u-1=0\end{align}$$
This leads to:
$$\begin{align}&u=\sqrt{1-x^2}-x\\
\implies &ux=u-x^2,\,x≠0\\
\implies &x^2+ux-u=0.\end{align}$$
Finally, note that all possible roots of the original equation must satisfy the inequality $0≤x≤\frac {\sqrt 2}{2}.$
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