Solving two equations in polar coordinates Problem:
Find the points of intersection of the following two curves.
\begin{align*}
r &= 1 - \cos \theta \\
r^2 &= \cos \theta
\end{align*}

Answer:
Note: While I do not show a plot, I did produce one. It indicated that there were only two solutions.
Now we solve for the points of intersection:
\begin{align*}
r^2 &= r^2 \\
\left( 1 - \cos \theta \right) ^2 &= \cos \theta \\
\left( \cos \theta - 1 \right) ^2 &= \cos \theta \\
\cos^2 \theta - 2 \cos \theta + 1 &= \cos \theta \\
\cos^2 \theta - 3 \cos \theta + 1 &= 0
\end{align*}
\begin{align*}
\cos \theta &= \dfrac{ 3 \pm \sqrt{ 9 - 4(1)(1) } } { 2(1) }
 = \dfrac{ 3 \pm \sqrt{ 5 } } { 2 }
\end{align*}
Here are the two values for $\theta$. Notice they are in the first and
fourth quadrant.
\begin{align*}
\theta &= \cos^{-1} \left(  \dfrac{ 3 + \sqrt{ 5 } } { 2 } \right) \\
\theta &= \cos^{-1} \left(  \dfrac{ 3 - \sqrt{ 5 } } { 2 } \right)
\end{align*}
Since $\cos(-x) = \cos(x)$ it would appear to me that there are four valid
solutions. However, the graph indicates that there are only two valid solutions. Did I do something wrong?

The book's answer is:
$$ \left( \dfrac{-1 + \sqrt{5}}{2},
 \cos^{-1}\left( \dfrac{3 - \sqrt{5}}{2} \right)  \right),
 \left( \dfrac{-1 + \sqrt{5}}{2},
  2\pi - \cos^{-1}\left( \dfrac{3 - \sqrt{5}}{2} \right)  \right) $$
I am  feeling confused and I am hoping somebody can shed some light
on the problem.

There are indeed only two valid solutions (up to $2\pi$ periodicity). Note that $|\cos(\theta)|\leq 1$ so we discard the case $\cos(\theta)=\frac{3+\sqrt{5}}{2}$. Then $\cos(-x)=\cos(x)$ does not introduce new solutions, since we have
$$\cos(\theta)=\frac{3-\sqrt{5}}{2}\implies\theta=\pm\cos^{-1}\left(\frac{3-\sqrt{5}}{2}\right)+2n\pi,\space n\in\mathbb{Z}$$
$$\cos(-\theta)=\frac{3-\sqrt{5}}{2}\implies\theta=\mp\cos^{-1}\left(\frac{3-\sqrt{5}}{2}\right)+2k\pi,\space k\in\mathbb{Z}$$
and they are the same set of solutions.
