Assume we want to know that where do the two polar curves $r=2\cos\left(\theta\right)$ and $r=2\sin\left(\theta\right)$ intersect each other.
What I know is that the intersection point is where the two curves have a common value,what I do in Cartesian coordinate is that I let the two curves to be equal and then solve for the determined variable.
However it looks that is totally different in polar coordinates,besides some elementary books like Thomas Calculus (12th) suggest some ways to find where do the polar curves intersect,unfortunately these ways are all incomplete and I'm totally sure that there must be more precious and more advanced way for that.
Can someone explain why equating the two curves does not give us the whole answers?
Besdies I really want to know what is an standard way to find the intersection point of two curves.
I'd say it's the same principle:
$$r=2\sin\theta=2\cos\theta\iff\sin\theta=\cos\theta\iff\theta=\frac\pi4\;\text{ or }\;r=0$$
since it must be $\;0\le\theta\le\frac\pi2\;$ as $\;r\ge0\;$ , and then $\;r=2\sin\frac\pi4=2\cos\frac\pi4=\sqrt2\;$ or $\;r=0\;$ , and you get both points of intersection.
Or passing to cartesian coordinates:
$$r=\sqrt{x^2+y^2}\;,\;\begin{cases}\sin\theta=\sin\arcsin\frac yx=\cfrac{\frac yx}{\sqrt{1+\frac{y^2}{x^2}}}=\cfrac y{\sqrt{x^2+y^2}}\\{}\\\cos\theta=\cos\arctan\frac yx=\cfrac1{\sqrt{1+\frac{y^2}{x^2}}}=\cfrac x{\sqrt{x^2+y^2}}\end{cases}$$
Then
$$2\cfrac y{\sqrt{x^2+y^2}}=2\cfrac x{\sqrt{x^2+y^2}}\iff x=y\;,$$
this time, only for $\;x,\,y>0\;$, plus the trivial equality with $\;x=y=0\;$