6. The smallest natural number $n$ that satisfies $n \sin 1 > 1 + 5 \cos 1$ is $\qquad$ .
6.5 .

Since $\frac{\pi}{4}n \sin 1>1+5 \cos 1>1+5 \cos \frac{\pi}{3}=\frac{7}{2} \text {. }$
Thus, $n>\frac{7}{\sqrt{3}}>4$.
When $n=5$, it is easy to prove that $5 \sin 1>1+5 \cos 1$, which means
$$
5(\sin 1-\cos 1)>1 \text {, }
$$

Squaring it gives $\sin 2<\frac{24}{25}$, which is obviously true.
In fact, because $\frac{3 \pi}{10}<1<\frac{\pi}{3}$, so,
$$
\begin{array}{l}
\sin \frac{2 \pi}{3}<\sin 2<\sin \frac{3 \pi}{5}=\frac{\sqrt{10+2 \sqrt{5}}}{4} \\
<\frac{\sqrt{14.5}}{4}<\frac{24}{25} .
\end{array}
$$

Therefore, the smallest natural number $n=5$.