How to find the number of intersection for $ \rho =\frac{\theta} {2\pi+1} $ and $\rho =\frac {1} {2-\cos\theta} $

Question

How to Find the number of intersection for curve $ \rho =\frac{\theta} {2\pi+1} $ and curve $\rho =\frac {1} {2-\cos\theta} $ .

Answer 1

$\cos{\theta}=2-\dfrac{2\pi+1}{\theta}, 1 \ge \cos{\theta} \ge -1 \to 1 \ge 2-\dfrac{2\pi+1}{\theta} \ge -1 \iff 3 \ge \dfrac{2\pi+1}{\theta} \ge 1 \iff 2\pi+1 \ge \theta \ge \dfrac{2\pi+1}{3}$

$\dfrac{\pi}{4} < 1 < \dfrac{\pi}{3} \to 2\pi + \dfrac{\pi}{3} > \theta > \dfrac{2\pi}{3}+\dfrac{\pi}{12}=\dfrac{3\pi}{4}$, now divide the domain into 4 segments:

$[\dfrac{3\pi}{4},\pi] ;[\pi,\dfrac{3\pi}{2}],[\dfrac{3\pi}{2},2\pi],[2\pi,2\pi + \dfrac{\pi}{3}]$

let $g(\theta)=2-\dfrac{2\pi+1}{\theta}$

in $[\dfrac{3\pi}{4},\pi], \cos{\theta} $ from $-\dfrac{\sqrt{2}}{2}$ to $-1,g(\theta)$ from $-1$ to $-\dfrac{1}{\pi}$ , they will corss each.(explain yourself)

in $[\pi,\dfrac{3\pi}{2}], \cos{\theta} $ from $-1$ to $0,g(\theta) $ from $-\dfrac{1}{\pi}$ to $\dfrac{2(\pi-1)}{3\pi}$,what happen to $ \cos{\theta},g(\theta)$.

I think you can do rest now.

Edit: I add two graphics to show the result:

Answer 2

$\theta $ can be negative

four intersections.