Convert $ x^2 - y^2 -2x = 0$ to polar? So far I got
$$r^2(\cos^2{\phi} - \sin^2{\phi}) -2 r\cos{\phi} = 0$$
$$r^2 \cos{(2\phi)} -2 r \cos{\phi} = 0$$

alright you are on a good track so from where you left of 
$r^2 cos(2\theta) - 2rcos\theta = 0 \Rightarrow r^2 cos(2\theta)= 2rcos\theta \Rightarrow
r cos(2\theta) = 2cos\theta \Rightarrow r = \frac{2cos\theta}{cos(2\theta}$
