Polar to cartesian form of $r=\cos(4θ)$? 
Consider an equation in polar coordinates, $r = \cos(4θ)$. Find the equation of the curve in the first quadrant in Cartesian coordinates.

This is for an assignment and this is what help I have received so far from user170231-
$$\begin{align}
r(\theta)&=\cos(4\theta)\\[1ex]
&=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\[1ex]
&=\frac{r^4\left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)}{r^4}\\[1ex]
&=\frac{(r\cos \theta)^4-6(r\cos \theta)^2(r\sin \theta)^2+(r\sin \theta)^4}{\left(r^2\right)^2}
\end{align}$$
He told me to see if I could solve it now, and here's what I could come up with:
$$\sqrt{x^2+y^2} = \frac{x^4-6x^2y^2+y^4}{(x^2+y^2)^2}$$
Is this step correct? If so, am I done with the transformation of $r=\cos(4θ)$ to cartesian for the first quadrant only? If not, where should I go from here? Any help is appreciated.

Welcome to MSE! Here you can see how $r=cos(4\theta)$ is negative and positive and the importance of solving for both $\pm \sqrt{x^2+y^2}$



