Solution

Note that, first we need to check if the graph is symmetric with respect to the polar axis, the line $\theta=\frac{\pi}{2}$ and the pole or not.

The polar axis:

We need to change $\theta$ to $-\theta$.

Hence $r=2 cos (-2 \theta)$ and since $cos (-2 \theta)= cos (2 \theta)$ we have $r= cos(2 \theta)$. So the graph is symmetric with respect to the polar axis.

The line $\theta=\frac{\pi}{2}$:

We need to plug in $\pi - \theta $ to the equation $r=2cos(2\theta)$. Hence $r=2 cos (2(\pi-\theta))=2 cos (2\pi-2\theta)=2cos(2\theta)$.

So the graph is symmetric with respect to the line $\theta=\frac{\pi}{2}$

Since the graph is symmetric with respect to both polar axis and the line $\theta=\frac{\pi}{2}$, it must be symmetric with respect to the pole too.

So, it's enough to graph $r=2 \cos (2 \theta)$ in the interval $[0,\frac{\pi}{2}]$.

We can plug in $\theta=0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$,$\frac{\pi}{2}$.

Hence $ r(0)=2$, $r(\frac{\pi}{3})=-1$,$r(\frac{\pi}{6})=1$ ,$ r(\frac{\pi}{2})=-2$ and $r(\frac{\pi}{4})=0$.

Now, we should be able to graph the $r=2 \cos (2 \theta)$ in $[0, \frac{\pi}{2}]$.

and hence, we can easily draw the graph in the $[\frac{\pi}{2}, 2 \pi] $.

Please check the graphs in the attached file.

I forgot to write theta is between 0 and 2*pi