How to find $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$? 
Let $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=f(x)$, where $f(u)=c\sin u$.Find $c$.

Trial:$\sum_{k=1}^{\infty}\frac{c \sin (x+k\pi)}{2^k}=c\sin x$.Then $c=0$ is a solution. Is there any other solution of $c$? Mainly I am interested in the sum $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$. please help.

Using a little trigonometry and the sum of a convergent infinite geometric series:
$$\sin(x+k\pi)=\sin x\cos k\pi+\sin k\pi\cos x=\sin x\cos k\pi=(-1)^k\sin x\Longrightarrow$$
$$c\sin x=\sum_{k=1}^\infty\frac{c\sin(x+k\pi)}{2^k}=c\sin x\sum_{k=1}^\infty\left(-\frac{1}{2}\right)^k=c\sin x\frac{-\frac{1}{2}}{1+\frac{1}{2}}=$$
$$=-\frac{c\sin x}{3}\Longrightarrow c\sin x=-\frac{c\sin x}{3}\Longleftrightarrow c=0\,\,\,\vee\,\,\,\sin x=0\Longleftrightarrow$$
$$\Longleftrightarrow c=0\,\,\,\vee\,\,\,x=n\pi\,\,\,,n\in\Bbb Z$$
