6. The smallest positive integer $n$ that satisfies $n \sin 1 > 5 \cos 1 + 1$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
6. (B).

From $n \sin 1>1+5 \cos 1$ we get $n \sin \frac{\pi}{3}>n \sin 1>1+5 \cos 1>1+5 \cos \frac{\pi}{3}$, which means $n>\frac{2+5}{\sqrt{3}}=\frac{7}{3} \sqrt{3}>4$. Therefore, $n \geqslant 5$.

Next, we prove that $5 \sin 1>5 \cos 1+1$, which is equivalent to proving $\sin 1 - \cos 1 > \frac{1}{5}$, or equivalently, proving $\sin \left(1-\frac{\pi}{4}\right) > \frac{\sqrt{2}}{10}$.
As shown in the figure, by the geometric convexity of $y=\sin x$, we obtain the inequality
$$
\sin x > \frac{\frac{\sqrt{2}}{\frac{\pi}{4}}}{x}\left(0 < x < \frac{\pi}{4}\right) > \frac{2 \sqrt{2}}{\pi} \cdot x \left(0 < x < \frac{\pi}{4}\right) > \frac{2 \sqrt{2}}{\pi} \left(1-\frac{\pi}{4}\right) > \frac{\sqrt{2}}{10} \text{ (since } 0 < 1-\frac{\pi}{4} < \frac{\pi}{4} \text{)}, \text{ i.e., (1) holds.} \text{ Hence, the smallest positive integer } n = 5.