How do I convert $r = - 3\sin \theta $ to rectangular form?

Hint: To convert the given expression from polar form to rectangular form, we must know that ${r^2} = {x^2} + {y^2}$ where $x = r\cos \theta $ and $y = r\sin \theta $. From the given expression, we have to calculate such values in which we can substitute these equations in order to eliminate $\theta $ and convert it into rectangular form.

Complete step by step solution:
(i)
We are given,
$r = - 3\sin \theta $
Since we know that,
${r^2} = {x^2} + {y^2}$
And,
$x = r\cos \theta $ -[eq.1]
And,
$y = r\sin \theta $ -[eq.2]
We can also prove it by squaring and adding eq.1 and eq.2, we will get:
${x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta $
Taking the common term ${r^2}$ out, we will get:
${x^2} + {y^2} = {r^2}\left[ {{{\cos }^2}\theta + {{\sin }^2}\theta } \right]$
As we know the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$, it will become:
${x^2} + {y^2} = {r^2}$
(ii)
We can clearly see if we multiply both sides by $r$ in $r = - 3\sin \theta $, we will get ${r^2}$ in LHS.
So, multiplying $r$ on both the sides, we will get:
${r^2} = - 3r\sin \theta $
(iii)
Since, we want our expression in terms of $x$ and $y$, we will substitute ${r^2}$ as ${x^2} + {y^2}$
${x^2} + {y^2} = - 3r\sin \theta $
(iv)
Also, as we know that $y = r\sin \theta $, we can substitute $r\sin \theta $ as $y$ in the above expression. So, after the substitution, we will get:
${x^2} + {y^2} = - 3y$
Shifting $ - 3y$ in the LHS so that we get our expression in rectangular form, it will become:
${x^2} + {y^2} + 3y = 0$
Hence, the rectangular form of $r = - 3\sin \theta $ is ${x^2} + {y^2} + 3y = 0$.

Note: Our basic aim is to convert the expression in terms of $x$ and $y$ and to remove the variables $r$ and $\theta $ given in the question statement. The most important step here is to identify how we bring the expression given in the question in a form where the values of $x$ and $y$ can be substituted to obtain the rectangular form.