Why $r$ is equal to $1/\phi$ in the System of Equations I have this system

$$\begin{align*} & \cos\theta\left(r+\frac 1r\right)=-2\\ & r^2+\frac 1{r^2}+4\cos^2\theta=10\end{align*}$$

I solved for $r$ and $\cos\theta$ by first squaring the first equation$$r^2+\frac 1{r^2}=\frac 4{\cos^2\theta}-2$$So$$\frac 4{\cos^2\theta}+4\cos^2\theta=12\quad\implies\quad\cos\theta=\frac {\pm1\pm\sqrt5}2$$

Question: 
  However after this, I'm not sure which value of the cosine I should take. The result used is $\cos\theta=1/\phi$, but I'm not sure why the other three solutions were discarded. The full solutions are$$\cos\theta=\frac 1{\phi}\quad\qquad\quad r=\phi+\sqrt \phi$$where $\phi=\tfrac 12\left(1+\sqrt5\right)$.

For context, I'm working out this problem which was solved by Ron Gordon. I'm having a hard time figuring out why he chose $1/\phi$ and $\phi+\sqrt{\phi}$.

Starting from
$$ \frac{4}{\cos^2\theta} + 4\cos^2\theta - 12 = 0 $$
You get
$$ \cos^4\theta - 3\cos^2\theta + 1 = 0 $$
which gives
$$ \cos^2\theta = \frac{3\pm\sqrt{5}}{2} $$
However $\cos^2\theta \le 1$, so we must take the negative sign. Furthermore you want $\cos\theta < 0$ since
$$ \cos\theta \left(r + \frac{1}{r}\right) = -2 < 0 $$
This leaves
$$ \cos\theta = -\sqrt{\frac{3-\sqrt{5}}{2}} = \frac{1-\sqrt{5}}{2} = -\frac{1}{\varphi} $$
which means
$$ r + \frac{1}{r} = 2\varphi $$
or 
$$ r = \varphi \pm \sqrt{\varphi} $$
both roots are equally valid as they're reciprocals of each other, and $r$ and $r^{-1}$ are interchangeable.
