Problem 6. Suppose $T$ is the period of the function $y=f(x)$. From the condition, we get:

$$
\sin x=f(x)-0.5 f(x-\pi)=f(x-T)-0.5 f(x-T-\pi)=\sin (x-T)
$$

Since this equality holds for all $x$, the number $T$ is the period of the function $y=\sin x$, i.e., $T=2 \pi n$, where $n$ is some natural number. From the given equation, we get:

$$
\begin{gathered}
f(x)=0.5 f(x-\pi)+\sin x=0.5(0.5 f(x-2 \pi)-\sin x)+\sin x= \\
=0.5(0.5(f(x-3 \pi)+\sin x)-\sin x)+\sin x= \\
=\ldots=0.5(0.5(\ldots(0.5 f(x-2 \pi n)-\sin x)+\ldots+\sin x)-\sin x)+\sin x= \\
=0.5^{2 n} f(x)-\sin x\left(0.5^{2 n-1}-0.5^{2 n-2}+0.5^{2 n-3}-\ldots+0.5-1\right)
\end{gathered}
$$

Then $f(x)\left(1-0.5^{2 n}\right)=\sin x\left(1-0.5+0.5^{2}-\ldots-0.5^{2 n-1}\right)=\sin x \frac{1-0.5^{2 n}}{1+0.5}$.

Therefore, $f(x)=\frac{2}{3} \sin x$.
Answer: $f(x)=\frac{2}{3} \sin x$.