Answer
$8x^{2}+9y^{2}-6x-9=0$
Work Step by Step
Multiply $3-\cos\theta$ to both sides pf the equation to obtain:
\begin{align*}
(3-\cos\theta)(r)&=\frac{3}{3-\cos\theta} \cdot (3-\cos\theta)\\
\\3r-r\cos\theta&=3
\end{align*}
Since $r=\sqrt {x^{2}+y^{2}}$ and $r\cos\theta=x$, then the equation above is equivalent to
\begin{align*}
3\left(\sqrt {x^{2}+y^{2}}\right)-x&=3\\
3\left(\sqrt {x^{2}+y^{2}}\right)&=x+3\\
\left(3\left(\sqrt {x^{2}+y^{2}}\right)\right)^2&=(x+3)^2\\
9(x^{2}+y^{2})&=x^2+6x+9\\
9x^{2}+9y^{2}&=x^2+6x+9\\
9x^{2}+9y^{2}-(x^2+6x+9)&=0\\
9x^{2}+9y^{2}-x^2-6x-9&=0\\
8x^{2}+9y^{2}-6x-9&=0\\
\end{align*}
