Answer
$ x^{2}-x+y^{2}=0\quad \quad \text{or}\quad \quad\left(x-\frac{1}{2}\right)^{2}+y^{2}=\frac{1}{4}$
Work Step by Step
Multiply $r$ to both sides of the equation, to obtain
$r\cdot r=r\cdot \cos\theta$
$r^{2}=r\cos\theta$
Since $r^{2}=x^{2}+y^{2}$ and $r\cos\theta=x$, then,
\begin{align*}
r^2&=r\cos\theta\\
x^{2}+y^{2}&=x\\
x^2-x+y^2&=0\end{align*}
The equation above can be written in standard form of a circle by completing the square:\begin{align*}
\left(x^{2}-x+\frac{1}{4}\right)+y^{2}&=0+\frac{1}{4}\\
\left(x-\frac{1}{2}\right)^2+y^{2}&=\frac{1}{4}\\
\end{align*}
Thus, equivalent of the given equation in rectangular coordinates is:
$$\left(x-\frac{1}{2}\right)^{2}+y^{2}=\frac{1}{4} \quad \quad \text{or} \quad \quad x^2-x+y^2=0$$
