<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] I wanted to provide the trigonometric approach as a $3$rd answer, thinking it might be useful to some readers.

Substitute $x=\sin\theta$, where $0\leqslant \theta\leqslant \frac {\pi}{2}$ based on the fact $x\geqslant 0$, then we have:
$$\begin{align}&\sin\theta+\sin \theta\cos \theta=\cos \theta\\
\implies &\cos \theta-\sin \theta=\sin \theta\cos \theta\\
\implies &1-\sin (2\theta)=\frac{\sin ^2 (2\theta)}{4}\end{align}$$
Since $\sin (2\theta)\geqslant 0$ we obtain:
$$\begin{align}&\sin^2(2\theta)+4\sin (2\theta)-4=0\\
\implies &\sin (2\theta)=2\sqrt 2-2\\
\implies &\theta=\frac 12\arcsin \left(2\sqrt 2-2\right)\end{align}$$
Finally, the answer is:
$$\bbox[5px,border:2px solid #C0A000]{x=\sin\left(\frac 12\arcsin \left(2\sqrt 2-2\right)\right)}$$

Important note:
The final answer assume that, for $0\leqslant \theta\leqslant \frac {\pi}{2}$ the inequality $\cos\theta\geqslant \sin\theta$ holds to avoid extraneous roots. No other root satisfies the inequality $\cos\theta\geqslant \sin\theta$, under the restriction $0\leqslant \theta\leqslant \frac {\pi}{2}.$
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