Answer
$(x^{2}+y^{2})^{\frac{3}{2}}-x=0$
Work Step by Step
$r^{2}=\cos\theta$
Multiply $r$ to both sides of the equation to obtain
\begin{align*}
r^{2} \cdot r&=r\cdot \cos \theta
\\r^3&=r\cos\theta\end{align*}
Since $r=\sqrt {x^{2}+y^{2}},\,\,r^{2}=x^{2}+y^{2}$ and $r\cos\theta=x$, then
\begin{align*}
r^3&=r\cos\theta\\
\left(\sqrt{x^2+y^2}\right)^3&=x\\
\left((x^2+y^2)^{\frac{1}{2}}\right)^3&=x\\
\left(x^2+y^2\right)^{\frac{3}{2}}&=x\\
\left(x^2+y^2\right)^{\frac{3}{2}}-x&=0
\end{align*}
