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.

$$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$
$$x^2\left(1+2\sqrt {1-x^2}+1-x^2\right)=1-x^2$$
$$x^2\left(2+2\sqrt {1-x^2}-x^2\right)=1-x^2$$
Substitute $x^2=y$,
$$y\left(2+2\sqrt {1-y}-y\right)=1-y$$
$$2y+2y\sqrt {1-y}-y^2=1-y$$
$$2y\sqrt {1-y}=1-y+y^2-2y$$
$$4y^2(1-y)=(y^2-3y+1)^2$$
$$y^4-2y^3+7y^2-6y+1=0$$
$$(y^2-y+3)^2=8$$
$$y^2-y+3=\pm 2\sqrt 2$$
$$y = \frac {1}{2}\bigg(1 \pm \sqrt{8 \sqrt 2 - 11}\bigg)$$
$$x=\pm\sqrt {\frac {1}{2}\bigg(1 \pm \sqrt{8 \sqrt 2 - 11}\bigg)}$$
Hope this helps.
