$$r = 4\cos\theta+1$$ $$r = 2\cos\theta+1$$
This system has simultaneous solutions at $(1, \frac\pi2)$ and $(1, \frac{3\pi}2)$. But looking at the graph, there are non-simultaneous intersections at $(3, 0)$ and the pole. How can that be determined algebraically? What is the general way of solving for non-simultaneous intersections? Thanks in advance.
Edit: It seems like if you have $f(\theta)$ and $g(\theta)$ you need to solve for $f(\theta) = g(\theta)$ as well as $f(\theta) = -g(\theta+\pi)$. Correct me if I am wrong.
In polar coordinate functions, replacing $(r,\theta)$ by $(-r,\pi+\theta)$ does not make any difference.
So in this example, $$r=4\cos \theta+1$$ $$-r=-2\cos \theta+1$$ $$\text{Then, }0=2\cos \theta+2$$ $$\theta = \pi\Rightarrow r=-3$$ Therefore $(-3,\pi)$(or $(3,0)$) are solutions.
(Please someone verify this)