Solving an Equation for y in terms of x How is it possible to solve the equation:
$$ x = \frac{2-y}{y} \sqrt{1-y^2} $$
for $ y $ in terms of $ x $?
I know that it is possible to do it, since WolframAlpha gives a complicated answer but entirely composed of elementary functions. However, I am stumped on how to even start.
Thanks in advance!
 A: The typical method would be to square both sides (possibly picking up extraneous solutions) and then arrange the terms into a polynomial equation in $y$:
$$x = \frac{2-y}{y} \sqrt{1-y^2}$$
$$x^2 = \frac{(2-y)^2}{y^2} (1-y^2)$$
$$y^2 x^2 = (4-4y+y^2)(1-y^2) = (4-4y+y^2) - 4y^2 + 4y^3 - y^4$$
Which yields:
$$y^4 - 4y^3 + (x^2 + 3)y^2 + 4y - 4 = 0$$
If this were simply a quadratic in $y$, you could use the quadratic equation. As it stands this is a fourth order polynomial. There is a procedure to find the roots of such a polynomial, and you can find it through the wiki page on Quartic Polynomials. It's a bit of a mess though.
A: for the $\sqrt{1-y^2}$ to make sense we need  $-1 \le y \le 1.$ therefore we can set $ y = \cos t.$ with that, we have $$x = \frac{2-\cos t}{\cos t}\sqrt{1- \cos^2 t} = \frac{(2 - \cos t)\sin t}{\cos t}, 0 \le t \le \pi.$$  we can solve $$\frac{(2 - \cos t)\sin t}{\cos t} = x \tag 1$$ for $$\sin t = \frac{x\cos t}{2 - \cos t}\to (2 - \cos t)^2\cos^2 t + x^2\cos^2t = 1 $$ that is $$
cos^4t-4\cos^3 t+(x^2+4)\cos^2t-1 = 0, y = \cos t.$$
A: This is not an answer but it is too long for a comment.
As shown in previous comments and answers, $$f(y) = \frac{2-y}{y} \sqrt{1-y^2}$$ is only defined for $-1\leq y\lt 0$ and $0\lt y\leq 1$. If $x>0$, for $f(y)=x$ ,there is only one solution corresponding to $y>0$ and if $x<0$ there is only one solution corresponding to $y<0$. All possible remaining roots   were introduced by squaring.
For a given large value of $x$, using Taylor series, $f(y)$ can be approximated by $$f(y)=\frac{2}{y}-1-y+O\left(y^2\right)$$ which give solutions $$y_{\pm}=\frac{1}{2} \left(-x-1\pm\sqrt{x^2+2 x+9}\right)$$ ($y_+$ corresponding to $x>0$ and $y_-$ corresponding to $x<0$).
For example, for $x=10$, the approximation gives $y\approx 0.178908$ while the exact solution should be $y\approx 0.179140$; for $x=-10$, the approximation gives $y\approx -0.216991$ while the exact solution should be $y\approx -0.216388$.
For a more rigorous numerical solution, Newton method could be used since we already have good initial values.
A better approximation of $f(y)$ is given by a Pade approximant $$f(y)\approx\frac{2-\frac{3 y^2}{2}}{y+\frac{y^2}{2}}$$ which would approximate the solution as $$y_{\pm}=\frac{-x\pm\sqrt{x^2+4 x+12}}{x+3}$$ For the above example, the aproximations are almost identical to the solutions.
