8. Given the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}$ satisfies: for any $x, y$ $\in \mathbf{R}_{+}$, we have
$$
f(x) f(y)=f(x y)+2 \cos \left(\frac{1}{x}+\frac{1}{y}+2000\right) \text {. }
$$

Then all functions $f$ that satisfy the condition are . $\qquad$
8. $f(x)=\frac{1}{x}+2006$.

Let $y=1$, we get
$$
f(x) f(1)=f(x)+2006\left(\frac{1}{x}+2006\right) \text {. }
$$

Let $x=1$, we get
$$
f^{2}(1)=f(1)+2006 \times 2007 \text {. }
$$

Therefore, $f(1)=-2006$ or $f(1)=2007$.
If $f(1)=-2006$, then from equation (1) we get
$$
f(x)=-\frac{2006}{2007}\left(\frac{1}{x}+2006\right) \text {. }
$$

If $f(1)=2007$, then from equation (1) we get
$$
f(x)=\frac{1}{x}+2006 .
$$

Upon verification, $f(x)=\frac{1}{x}+2006$ satisfies the given conditions.