Step 3 is not equivalent to step 1 however, according to desmos here when zoomed in (a lot).
With further insight from the comments section, I have learned Desmos can become inaccurate when zoomed in very (very) closely, which is what caused me to believe my answer was incorrect.
$\begingroup$ When I visit your Desmos link and zoom in, the two graphs seem identical to me. What makes you say they are different? $\endgroup$
Commented Jun 5, 2018 at 3:42$\begingroup$ It looks the same for me, but even if it looks wrong, desmos is very good at generating graphs, but it is still not perfect, above certain values and close enough to 0 there are going to be errors, if Iirc desmos will be accurate enough till around $10^$ and then start to have a bit errors $\endgroup$
Commented Jun 5, 2018 at 3:56 $\begingroup$ ah I see, desmos is inaccurate when zoomed in a lot. thank you $\endgroup$ Commented Jun 5, 2018 at 6:18What you tried is perfect, the two equations are almost equivalent, see e.g. the answer I gave here. In your case $$r^2=r\cos\theta+r\sin\theta\Longleftrightarrow\cases\\r=0>$$
So the only caveat is what happens when $r=0$ : the value of $\theta$ is undefined at the origin, so it does not seem clear whether or not the equation $r=\cos\theta+\sin\theta$ is satisfied.
However, you can look at it this way: the polar curve of equation $r=>$ is the set of all points $(>\cos\theta,>\sin\theta)$ when $\theta$ runs over $[0,2\pi]$ , and this includes the point $\theta=\frac<3\pi>$ where. $r=0$ .
From this point of view, one can argue that the equation $r=\cos\theta+\sin\theta$ is satisfied when $r=0$ as well.