# Polar equation of cartesian $y = 1 + 3x$

I have no idea at all what to do on this

I got $$\cos^{-1} \left(\frac{r\sin \theta+1}{3} \right) = \theta$$

Which can be

$$\cos^{-1} \left(\frac{r\sin \left(\cos^{-1} \left(\frac{r\sin \theta+1}{3} \right) \right)+1}{3} \right) = \theta$$

$$\cos^{-1} \left(\frac{r\sin \left(\cos^{-1} \left(\frac{r\sin \left(\cos^{-1} \left(\frac{r\sin \theta+1}{3} \right) \right)+1}{3} \right) \right)+1}{3} \right) = \theta$$

So that obviously seems wrong, other than that 30 minutes of work has produced nothing.

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All you need to do in problems like these is to just substitute $x= r \cos(\theta)$ and $y = r \sin(\theta)$ and write out the resulting equation. In this case, you get $$r \sin(\theta) = 1 + 3 r \cos(\theta)$$which on rearranging gives us $$r (\sin(\theta) - 3 \cos(\theta)) = 1 \implies r = \dfrac1{\sin(\theta) - 3 \cos(\theta)}$$ This is all we need to do to convert it into polar form.

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Is it also applicable for a Cartesian equation like this: x + y = 9? – John Ortiz Ordoñez May 11 '15 at 20:09