Solving a system of two equations in polar coordinates Below is a problem I did. I believe the answer is right. Is it? However, I am not sure my reasoning is correct. I am also interested in comments about my style.
Problem:
Find the points of intersection of the follow two pairs of curves.
\begin{align*}
r &= a(1 + \cos \theta) \\
r &= a( 1 - \sin \theta )
\end{align*}
Answer:
\begin{align*}
a \left(1 + \cos \theta \right) &= a \left( 1 - \sin \theta \right) \\
1 + \cos \theta &=  1 - \sin \theta  \\
\cos^2 \theta &=  \sin^2 \theta = 1 - \cos^2 \theta \\
2 \cos^2 \theta &= 1 \\
\cos \theta &= \pm \frac{1}{ \sqrt{2}}
\end{align*}
Consider $\theta = \frac{\pi}{4}$ as a solution. This corresponds to an $r$ value
of $a\left( 1 + \frac{ \sqrt{2}}{2} \right)$. However, this value does not satisfy the second
equation so it must discarded. Now, we consider $\theta = \frac{3\pi}{4}$. For the first equation, I get:
$$ r = a \left( 1 - \frac{1}{ \sqrt{2}} \right) $$.
For the second equation, I get:
$$ r = a \left(  1 - \frac{1}{ \sqrt{2}} \right)  $$
Hence, one of the points of intersection is:
$\left(  a\left( 1 + \frac{ \sqrt{2}}{2} \right), \frac{ 3 \pi }{4} \right)$
Now consider $\theta = \frac{5\pi}{4}$ as a solution. In this case, I have:
\begin{align*}
a(1 + \cos\left(  \frac{5\pi}{4} \right)  ) &= a( 1 - \frac{ \sqrt{2}}{2} ) \\
a(1 - \sin\left(  \frac{5\pi}{4} \right)  ) &= a( 1 - \frac{ \sqrt{2}}{2} ) 
\end{align*}
Hence, one of the points of intersection is:
$$ \left( a( 1 - \frac{ \sqrt{2}}{2} ) ,  \frac{5\pi}{4} \right) $$
Now, we need to consider $\theta =  - \frac{5\pi}{4} $
In this case, I have:
\begin{align*}
a(1 + \cos\left( - \frac{5\pi}{4} \right)  ) &= a( 1 - \frac{ \sqrt{2}}{2} ) \\
a(1 - \sin\left( - \frac{5\pi}{4} \right)  ) &= a( 1 + \frac{ \sqrt{2}}{2} ) 
\end{align*}
Hence  $\theta =  - \frac{5\pi}{4}$ is not a solution of the original equation. We picked this solution when we squared both sides.
Now I claim that $(0,a)$ and $(0,b)$ are the same point for all real numbers $a$ and $b$. Hence,
we have to consider the fact that both equations go through the orgin. Therefore, I claim that
$(0,0)$ is the third point of intersection even though $(0,0)$ is not on either curve.

In a Cartesian coordinate system two points $(x_1,y_1)$ and $(x_2, y_2)$ are the same only if $x_1=x_2$ and $y_1=y_2$. In the polar coordinate system, two points are identical if $r_1=r_2$ and either $\theta_1=\theta_2$ or $r_1=r_2=0$. The origin is special. Since both curves contain the origin, it is not necessary to have $\theta_1=\theta_2$ at that point. It is only for $r\ne0$ you solve the way you did.
You could have simplified a little bit your calculations. You know $$\cos\theta=-\sin\theta$$
You squared this, and you got two answers for $\cos\theta$, then you checked all four quadrants where $|\cos\theta|=|\sin\theta|=\frac 1{\sqrt 2}$. You could have immediately skipped first and third quadrant, where both sine and cosine have the same sign.
