Does the general solution with these four conditions exist? I want to find the functions which are satisfied the following conditions. Then the function $f(\theta)$ is defined on $[0 \ \ 2\pi ]$.
1,
When $f(\theta)=0$, $\cfrac{\partial f}{\partial \theta} \leq 0$
2,
$\displaystyle \int_{0}^{2 \pi} \biggl( \cfrac{\partial^{2} f}{\partial \theta^{2}} \biggr)d \theta=0$
3,
The equation $f(\theta)=0$ has three solutions. Particularly, $f(0)=0$ and $f(2\pi)=0$.
4,
$\min \{f(\theta)\}<0$ and $\max \{ f(\theta)\}>0$
It can be probably understood easier to consider like the diagram. Suppose to the polar coordinate system of the unit circle. For example, the function is defined $f(\theta)=\sin \theta$ that it can be described as an inclination of an ellipse like the figure. However, this case is not the right answer because first mention is unsatisfied. 

My opinion
I think that it can be satisfied all conditions to improve $\sin \theta$ like below figure. 

After I asked the question, I found the function which is one of the answer.
$$f(\theta)=\sin \theta ^{2n+1} \ \ \ \ (n=1,2, \cdots)$$
This graph is described as $n=3$.

Do you know else functions? When someone found another function, I will vote your answer. Additionally, is there the general solution? If someone discover it, I will accept your answer.

Well, apparently, the solutions are plenty. $-\sin\theta\cdot(1+\cos\theta)$ would do. Also, any linear combination of solutions is a solution too. So is any solution multiplied by any strictly positive smooth periodic function. And we haven't even started yet to consider the functions in which the second zero would be not at $\theta=\pi$ but someplace else...
No, I don't think we can reasonably describe the general solution.
Upd. I mean, we kinda can, but it would be in terms like "take any periodic function with two zeros per period, and transform it such and such..."
