Converting Polar Equation to Cartesian Equation problem So I have
1. $$\frac{r}{3\tan \theta} = \sin \theta$$
2. $$r=3\cos \theta$$
What would be the Cartesian equation???
 A: First note that 
$$\left\{ \matrix{
  x = r\cos \theta  \hfill \cr 
  y = r\sin \theta  \hfill \cr}  \right.$$
the general approach will be to solve for $r$ and $\theta$ and replace in your polar equation. However, in most times there are some shortcuts. See the following for the second one
$$\eqalign{
  & r = 3\cos \theta   \cr 
  & r = 3{x \over r}  \cr 
  & x = {1 \over 3}{r^2}  \cr 
  & 3x = {x^2} + {y^2} \cr} $$
and hence your final equation will be
$${x^2} + {y^2} - 3x = 0$$
which is a conic section. Specifically, it is a circle. I leave the first one for you. :)
A: You could substitute for $r$ and $\theta$ their expressions in terms of $x$ and $y$. But remember that trigonometric functions of $\theta$ have easier expressions than $\theta$ itself.
A: the first
$$\frac{r}{3\tan \theta} = \sin \theta$$
$$\frac{r\cos \theta}{3\sin \theta} = \sin \theta$$
$$\frac{x}{3\sin \theta} =\sin \theta$$
$$x =3\sin^2 \theta$$
$$r^2x =3r^2\sin^2 \theta$$
$$x(x^2+y^2)=3y^2$$
The second
$$r=3\cos \theta$$
$$r^2=3r\cos \theta$$
$$r^2=3x$$
$$x^2+y^2=3x$$
