<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] $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$
substitute $x=\sqrt{y}$
$$\sqrt y\left(1+\sqrt{1-y}\right)=\sqrt{1-y}$$
divide both sides by $\sqrt{y}$
$$1+\sqrt{1-y}=\sqrt{y^{-1}-1}$$
square both sides
$$2-y+2\sqrt{1-y}=y^{-1}-1$$
isolate the square root
$$2\sqrt{1-y}=y+y^{-1}-3$$
square both sides
$$4-4y=y^2+y^{-2}-6y-6y^{-1}+11$$
multiply both sides by $y^2$
$$4y^2-4y^3=y^4-6y^3+11y^2-6y+1$$
subtract the left hand side from the right
$$y^4-2y^3+7y^2-6y+1=0$$
Factoring $y^4-2y^3+7y^2-6y$ we get
$$y^4-2y^3+7y^2-6y=y(y^3-2y^2+7y-6)=y(y-1)(y^2-y+6)=(y^2-y)(y^2-y+6)$$
Substitute $y^2-y=z$
$$y^4-2y^3+7y^2-6y=z^2+6z$$
Plugging it back in we get
$$z^2+6z+1=0$$
$$(z+3)^2-8=0$$
$$z=-3±\sqrt{8}$$
Undoing the substitution we get
$$y^2-y+3±\sqrt{8}=0$$
$$y=\frac{-(-1)±\sqrt{(-1)^2-4(1)(3±\sqrt{8})}}{2(1)}=\frac{1±\sqrt{-11±4\sqrt{8}}}{2}$$
Knowing $x=\sqrt{y}$ we get the following
$$x=\sqrt\frac{1±\sqrt{-11±4\sqrt{8}}}{2}$$
Three of these solutions are extraneous
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