So, I've never formally studied polar graphs or their properties and/or behaviour, but I came across a question which assumes some of this knowledge and I'm a little curious as to how to "rigorously" arrive at the desired answer.
Let $g$ and $f$ be functions in the polar plane such that $g(\theta)=4\sin(\theta)$ and $f(\theta)=4\sin(2\theta)$. Hence, we have a polar graph looking like
where $g$ is the function that maps $\theta$ to the purple circle, and $f$ is the function that maps $\theta$ to the 4-petal flower. Now, by inspection, its pretty obvious that there are intersections at $\pi/3$ and $2\pi/3$, but when I went about showing that these are the points of intersection I ran into an issue (of sorts).
The intersections occur for values of $\theta$ such that $g(\theta)=f(\theta)$, hence we consider:
$$4\sin(\theta)=4\sin(2\theta)$$ $$\sin(\theta)=2\sin(\theta)\cos(\theta)$$ $$2\sin(\theta)\cos(\theta)-\sin(\theta)=0$$ $$\sin(\theta)(2\cos(\theta)-1)=0$$
Given that by inspection we can tell $\theta=\arcsin(0)$ yields no useful solutions, we consider when
$$\theta=\arccos(\frac{1}{2})$$ $$\implies \theta_1=\frac{\pi}{3}$$
Now, obviously the next value of $\theta$ using the equation above will be $2\pi-\pi/3$, but given that both intersections occur for $\theta \in[0,\pi]$, we can safely ignore this value. Now the issue is that there are two intersections, yet we have only found one. Given that the second intersection occurs in the second quadrant, I simply stated that "As the second intersection occurs in the second quadrant, where $\cos(\theta)<0$, consider when $\theta=\arccos(-1/2)$", which does yield the desired solution, but it doesn't act in accordance with the very original equation, which requires $\sin(\theta)(2\cos(\theta)-1)$ to equal $0$. Can anyone explain the situation to me so that I'll know how to go about these problems in future?
Any help is appreciated, thank you.
Your method is perfectly correct, and it's the graph that is throwing sand into your eyes. Apparently, <your_graphing_calculator> is doing some manipulation to get a nice looking flower :D
Note that there exist unique polar coordinates only when $f(\theta)=r\geq0$. Hence, your $f(\theta)$ graph (red petals) should only exist for $\theta\in[0,\pi/2]\cup[\pi,3\pi/2]$ i.e. for $\sin(2\theta)\geq0$. By this, it is now clear that there is only one intersection (in the first quadrant), as you yourself found out.
PS: Interestingly enough, this does apparently not apply to $g(\theta)=4\sin\theta$. Reason being that for $\theta\in[\pi,2\pi]$, $r\leq0$, so the "head of the arrow" (or whatever you may want to call it) is pointing again above the axes, so the circle traces itself back again. On a related point, note that when $\theta\in[\pi/2,\pi]$, $f(\theta)$ is actually drawing the petal in the fourth (not second!) quadrant, and when $\theta\in[3\pi/2,2\pi]$, the second (not fourth!) quadrant, since in both cases $r\leq0$, $r$ and $\theta$ are in opposite quadrants.