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.

Another variation.
We observe that, $\sqrt {1-x^2}≥0$ and $1+\sqrt {1-x^2}>0$ imply $x≥0$.
Letting $\sqrt {1-x^2}=u,~u≥0$ we have:
$$\begin{cases}
ux=u-x\\u^2+x^2=1\end{cases}$$
Then setting $u=a+b,~ x=a-b$ under the restriction $a≥b≥0$, we have:
$$\begin{align}&\begin{cases}a^2-b^2=2b\\a^2+b^2=\frac 12\end{cases}\\
\implies &4b^2+4b-1=0\\
\implies &x=\frac{\sqrt {4b+1}}{2}-b.\end{align}$$
This completes the answer.
