Find periodicity of a function $f(x)=\frac{\sin x}{2+\cos x}$

Question

$$\exists p\neq 0 : x\in D (f(x+p)=f(x))$$ where $D$ is the domain of $f(x)$.

$$f(x+p)=\frac{\sin (x+p)}{2+\cos (x+p)}=\frac{\sin p\cos x+\sin x\cos p}{2+\cos p\cos x-\sin x\sin p}$$

$$\frac{\sin x}{2+\cos x}=\frac{\sin p\cos x+\sin x\cos p}{2+\cos p\cos x-\sin x\sin p} /(2+\cos x)(2+\cos p\cos x-\sin x\sin p)$$

$$\sin x(2+\cos p\cos x-\sin x\sin p)=(\sin p\cos x+\sin x\cos p)(2+\cos x)$$

How to reduce this equation to find $p$?

Answer 1

Let's set a equation \begin{equation} \frac{\sin x}{2+\cos x} = \frac{\sin (x+p)}{2+\cos(x+p)}. \end{equation} It can be manipulated as \begin{equation} 2\sin x +\sin x \cos (x+p) = 2\sin(x+p)+\cos x \sin(x+p). \end{equation} By trigonometric identities, \begin{equation} \sin(x+p)-\sin x =2\cos\left(x+\frac{p}{2}\right)\sin\frac{p}{2}, \end{equation} \begin{equation} \sin(x+p)\cos x -\sin x \cos(x+p) =\sin(x+p-x)=\sin p, \end{equation} and \begin{equation} \sin p =2\sin\frac{p}{2}\cos\frac{p}{2}. \end{equation} Therefore, a equation reduces to \begin{equation} 4\cos\left(x+\frac{p}{2}\right)\sin\frac{p}{2}+2\sin \frac{p}{2}\cos\frac{p}{2}=0, \end{equation} as same as \begin{equation} \sin\frac{p}{2}\left(2\cos\left(x+\frac{p}{2}\right)+\cos\frac{p}{2}\right)=0. \end{equation} This equation must be true for all $x$. Let $x=-\frac{p}{2}$, then it reduces to \begin{equation} \sin\frac{p}{2}=0 \end{equation} and solution is $p=\pm 2\pi, \pm 4\pi,\dots$ Therefore period of $f(x)$ is $2\pi$.

Answer 2

The given function obviously has period $2\pi$, but there could be a smaller period, which then has to be a natural fraction of $2\pi$. As $f(x)=0$ exactly when $x\in \pi{\mathbb Z}$ we only have to test whether $\pi$ is a period. Since $f\bigl(-{\pi\over2}\bigr)=-1\ne1=f\bigl({\pi\over2}\bigr)$ this is not the case.