How to find the domain $\theta$ for each petal in the Polar graph $r=4\cos(3\theta)$? Given the equation $r=4\cos(3\theta)$, how can I find the domain of each petal? Help!
 A: This is the polar graph of your function $r=4\cos(3\theta)$:

As mentioned in the comment from @coffeemath; by domain I will assume that you mean the values for $\theta$ to make each petal. So the interval in this case is simply $\cfrac{2\pi}{3}$ for each petal as the full domain ($2\pi$) must be divided equally across each petal. So this means that $-\cfrac{\pi}{6} \le \theta \le \cfrac{\pi}{6}$ is the domain for the first petal (centred on the $x$-axis).
A: We might rephrase your question as follows: "Which values of $\theta$ correspond to where  (1) $r(\theta) = 0$ and (2) the function $r(\theta)$ has not yet become repetitious?"
Ok, so we understand that trigonometric functions are periodic. So what I mean to say is not only are we seeking $\theta$'s at which $r(\theta) = 0$ (which as you should know corresponds to the "shutting closed" of individual petals), but we are also seeking that these $\theta's$ belong to some closed interval where $r(\theta)$ has only repeated itself once. If you think about it, this will naturally correspond to the first petal in our sequence of repeating petals on the rose curve (it is the first repetition or cycle out of many little cycles making up the whole curve).
When considering these non-repetitious $\theta$'s, we also have to define what we mean by "repetitious". We can consider the interval where our trig function has cycled through both negative and positive $r(\theta)$ values once – or maybe we can consider where it has cycled through only the positive $r(\theta)$ values exclusively?  We should do the second option here because the first petal we know has positive $r$ values (you should know, however, that if you had $-4$ instead of $4$ you'd be doing the opposite because the first petal would have a negative $r$).
Look at the Cartesian graph of the parent cosine function. I want you think about how the values in $\large [-\frac{\pi}{2}, \frac{\pi}{2}]$ are related to $\cos$'s polar representation. 

When we solve your equation $4\cos 3\theta = 0$, we will arrive at a point where we have $3\theta = \cos ^{-1} 0$. Because we understand that $\cos$ is periodic, we now ought to consider the set of all $\theta$'s our inverse cosine can return. There are infinitely many.* However, we are safe to restrict our interest to the interval $[-2\pi, 2\pi]$ because beyond this interval we might say the standard parent trig functions have exceeded a sort of "max repetitiousness."**  
And so the way I like to do this algebraically is to understand $\cos ^1$ as giving me back a set of $\theta$'s between $[-2\pi, 2\pi]$. 
So I write this out $3\theta = \large \{ -\frac{3\pi}{2} ,-\frac{\pi}{2},\frac{\pi}{2}, \frac{3\pi}{2} \}$.
The way I like to do this algebraically is to consider $\cos ^1$ as giving me back a set of $\theta$'s between $[-2\pi, 2\pi]$. So I write this out $3\theta = \large \{ -\frac{3\pi}{2} ,-\frac{\pi}{2},\frac{\pi}{2}, \frac{3\pi}{2} \}$. Like I discussed earlier, I am going only to consider $ \large -\frac{\pi}{2}$ and $\large \frac{\pi}{2}$ because between these two values of $\theta$ we evaluate that $r(\theta)$ is positive and corresponds to the first positive $r$-valued petal. We divide the set $\large \{ -\frac{\pi}{2}, \frac{\pi}{2} \}$ now by $3$ finally in order to get $\large \{ -\frac{\pi}{6}, \frac{\pi}{6}\}$.
Your answer should be $\large \{ -\frac{\pi}{6}, \frac{\pi}{6}\}$ as the domain of the first petal. The solution is also illustrated graphically below.*** I hope this helps!

* Notwithstanding that, as you may already know, the inverse trig functions are often defined as having restricted domains so they behave as functions rather than relations.
** You should already know that every periodic solution for $\theta$ can be written of the form $k \pm 2\pi n$, where $k$ is some real number principal angle and $n$ is a natural number corresponding to a variable number of rotations. There's many different ways of conceptually appreciating why this is the case that I won't get into here.
*** The animated graph can be found here https://www.desmos.com/calculator/cakyyu0qgr. It's interactive and I highly suggest you take a look at it. 
