<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] Let us make the substitution $y := \sqrt{1-x^2}$; then we get the system of polynomial equations
\begin{align*}
x^2 + y^2 & = 1, \\
x(1+y) & = y.
\end{align*}
From here, if we solve the second equation for $y$, $y = \frac{x}{1-x}$, and substitute into the first equation, it is straightforward to get the polynomial equation
$$x^4 - 2x^3  + x^2 + 2x - 1 = 0.$$
However, let us go back to the original system of two equations in $(x, y)$.  Here, the first equation is the unit circle, and the second equation is equivalent to $(x-1)(y+1) = -1$, so it is a hyperbola with center $(1, -1)$.  We see that both curves are symmetric about the line $y = -x$; therefore, if $(x, y)$ is a solution to the system of equations, then so is $(-y, -x)$.  This implies that if $\alpha$ is a root of the quartic polynomial, then so is $-\frac{\alpha}{1 - \alpha} = \frac{\alpha}{\alpha - 1}$.
From here, we can hypothesize that if $K$ is the splitting field of the quartic over $\mathbb{Q}$, then there might be an element $\sigma \in \operatorname{Gal}(K / \mathbb{Q})$ which permutes the roots according to $\alpha \mapsto \frac{\alpha}{\alpha - 1}$.  If this is the case, then it is easy to see that $\sigma^2 = \operatorname{id}$ since it fixes each root of the quartic; however $\sigma \ne \operatorname{id}$ since the only fixed point of $\alpha \mapsto \frac{\alpha}{\alpha - 1}$ is 2, and 2 is not a root of the quartic.  Then setting $L$ to be the fixed field of $\sigma$, we will have $\operatorname{Tr}^K_L(\alpha) = \alpha + \frac{\alpha}{\alpha - 1} = \frac{\alpha^2}{\alpha - 1}$.  This suggests searching for a polynomial which will have $\frac{\alpha^2}{\alpha - 1}$ as a root, and hopefully finding the roots of this polynomial will be simpler because $L$ has smaller degree over $\mathbb{Q}$.
However, from this point, we fairly quickly see that
$$\left( \frac{\alpha^2}{\alpha-1} \right)^2 - 2 \left( \frac{\alpha^2}{\alpha-1} \right) - 1 = \frac{\alpha^4 - 2\alpha^2(\alpha-1) - (\alpha-1)^2}{(\alpha-1)^2} = \\
\frac{\alpha^4 - 2\alpha^3 + \alpha^2 + 2\alpha - 1}{(\alpha-1)^2} = 0.$$
But then, that means that $\frac{\alpha^2}{\alpha-1} = 1 \pm \sqrt{2}$.  Another application of the quadratic formula to the equation $\alpha^2 - (1 \pm \sqrt{2}) \alpha + (1\pm\sqrt{2}) = 0$ gives the roots of the quartic.  (Then from here, two of the roots of the quartic will be complex, and one of the real roots will not satisfy the original equation because it does not satisfy $\sqrt{1-x^2} = \frac{x}{1-x} \ge 0$.)
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