Convert $ x^2 - y^2 -2x = 0$ to polar?

Question

So far I got

$$r^2(\cos^2{\phi} - \sin^2{\phi}) -2 r\cos{\phi} = 0$$

$$r^2 \cos{(2\phi)} -2 r \cos{\phi} = 0$$

Answer 1

You are on the right track. Now divide through by $r \ne 0$ and get

$$r \cos{2 \phi} - 2 \cos{\phi} = 0$$

or

$$r = 2 \frac{ \cos{\phi}}{\cos{2 \phi}}$$

Answer 2

alright you are on a good track so from where you left of

$r^2 cos(2\theta) - 2rcos\theta = 0 \Rightarrow r^2 cos(2\theta)= 2rcos\theta \Rightarrow r cos(2\theta) = 2cos\theta \Rightarrow r = \frac{2cos\theta}{cos(2\theta}$