How does one correctly solve for the graph of $r^{2} = 36sin(2θ) $?

Question

What I did to solve was:

$r^{2} = 36sin(2θ)$

$r = \sqrt{36sin(2θ)}$

$r = 6 \sqrt{sin(2θ)}$

I set the $2θ = π/2$

so the incriminator is $π/4$

So by incrementing π/4:

$θ = 0, π/4, π/2, 3π/4, π/ 5π/4, 3π/2, 7π/4$

And the corresponding r’s evaluated are:

$r = 0, 6, 0, 6i, 0, 6, 0, 6$

So the pairs are: $0,0$

$π/4, 6$

$π/2, 0$

$3π/4, 6i$

$π, 0$

$5π/4, 6$

$3π/2, 0$

$7π/4, 6$

When I graph these I get a rose. What am I doing wrong? Thank you

Answer 1

Here is a simple plot of the function: