Using polar coordinates to find solutions to equation Using polar coodinates, find all the points $(x,y)\ne 0$ such that $\frac{2xy}{x^2+y^2}=\frac{1}{2}$.
I thought I could somehow substitute $x=r\cos(\theta), y=r\sin(\theta)$ but I'm not sure how to proceed. What strategy should I use to tackle these types of problems? Thanks!
EDIT: The numerator should be $2xy$
 A: The substitution is exactly what you should use, except that you don't have $x=r\cos (x)$, but rather $x=r\cos(\phi)$.
Then, using the fact that $\sin^2\phi + \cos^2\phi = 1$, the denominator on your right side should simplify quite a bit.
A: $x = r\cos\theta, y = r\sin\theta$
$\frac{2xy}{x^2+y^2} = \frac{2r^2\sin\theta\cos\theta}{r^2} = \sin 2\theta$
Solve $\sin 2\theta = \frac 12$ in the first four quadrants (only two quadrants are applicable).
Then, to get the solution in terms of $x$ and $y$, think about the equation(s) of line(s) subtending that angle with the horizontal axis. What is the gradient of those line(s)?
A: By your work $$r^2=4r^2\cos\theta,$$ which gives $$r=0,$$ which is impossible, or $$1=4\cos\theta\sin\theta,$$ which is $$\sin2\theta=\frac{1}{2}.$$
A: As $(x,y)\neq0$, you have $x=r\cos(\theta)$ and $y=r\sin(\theta)$ for $r>0$ and $\theta \in [0,2\pi]$.
Your equation become
$$\frac{2xy}{x^2+y^2}=\frac{2r^2 \cos(\theta)\sin(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))} = 2\cos(\theta)\sin(\theta)=\sin(2\theta)$$
So you get to solve
$$\sin(2\theta) = \frac{1}{2}$$
Which is now pretty simple with basic trigonometry knowledge.
