How can I find the points of intersection between the curves $r=1+\sin\theta$ and $r=1-\sin\theta$? Find the points of intersection for the curve $r=a(1+\sin\theta)$ and $r=a(1-\sin\theta)$
My book says the answer is $(0,0),(a,0),(a,\pi)$.
However I calculated $ (a,0),(a,\pi),(a,2\pi)$.
 A: Two of your points, $(a,0)$ and $(a,2\pi)$, are the same. 
It is easy to miss the solution $(0,0)$, since $(0,\theta)$ is the origin whatever $\theta$ is: the origin has many polar addresses.  
So $(0,0)$ is on one of the curves in the disguise $(0,\pi/2)$, and on the other in the disguise $(0,3\pi/2)$. 
To spot this sort of thing, one has to examine algebraically the possibility that $r=0$ for some $\theta$. Or make a sketch. 
A: For intersection, $1-\sin\theta=1+\sin\theta\implies \sin\theta=0\implies \theta=n\pi$ where $n$ is any integer
$\implies r=a(1\pm \sin n\pi)=a$
As the period of $\sin\theta,\cos\theta$ is $2\pi$
So, the in-congruent values of $\theta$ are $0,\pi$

Alternatively,
If $r=0,$ from the first equation $\sin\theta=-1$ and $\sin\theta=1$  from the second
$\implies r\ne0$
$r=a(1+\sin\theta)\implies r^2= ar+ ar\sin\theta\implies x^2+y^2-ar=ay$
Similarly, $r=a(1-\sin\theta)\implies x^2+y^2-ar=-ay$
$\implies ay=-ay\implies y=0\implies r=|x|, x^2-a|x|=0\implies |x|=a$ as $x=0$ would imply $ r=0$
So, $x=\pm a\implies r=a,\theta= \arctan \frac{0}{\pm a}=n\pi$ whose in-congruent values   are $0,\pi$
A: Suppose that $(r_0,\theta_0)$ is a solution to both $r=a(1+\sin(\theta))$ and $r=a(1-\sin(\theta))$.
That is, suppose
$$r_0=a(1+\sin(\theta_0))\quad\text{and}\quad r_0=a(1-\sin(\theta_0)).$$
If $a=0$, then we must have $r_0=0$ and $\theta_0$ can be anything (but all of these points are the same, namely, they all describe the origin; the origin doesn't have a well-defined angle).
Now suppose that $a\neq 0$. Then we can divide both sides by $a$ to see that
$$\begin{align*}
a(1+\sin(\theta_0))=a(1-\sin(\theta_0)) &\iff 1+\sin(\theta_0)=1-\sin(\theta_0)\\
&\iff 2\sin(\theta_0)=0\\
&\iff \sin(\theta_0)=0\\
&\iff  \theta_0\in\{0,\pi\}
\end{align*}$$
so that the solutions are
$$(a(1+\sin(0)),0)=(a,0)\qquad\text{ and }\qquad (a(1+\sin(\pi)),\pi)=(a,\pi).$$
