6. (10th "Hope Cup" Invitational Competition Question) Given $\sin \left(x+\arccos \frac{4}{5}\right)=\frac{\sqrt{3}}{2}, 0<x<\pi$, then $\sin x=$

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6. Fill $\frac{1}{10}\left(4 \sqrt{3}\right.$ and 3). Reason: From $0<x<\pi$ and $\sin \left(x+\arccos \frac{4}{5}\right)=\frac{\sqrt{3}}{2}$, we know that $x+\arccos \frac{4}{5}=\frac{\pi}{3}$ or $x+$ $\arccos \frac{4}{5}=\frac{2}{3} \pi$. Therefore, $\sin x=\sin \left(\frac{\pi}{3}-\arccos \frac{4}{5}\right)=\frac{\sqrt{3}}{2} \cdot \frac{4}{5}-\frac{1}{2} \cdot \frac{3}{5}=\frac{1}{10}(4 \sqrt{3}-3)$, or $\sin x=$ $\sin \left(\frac{2}{3} \pi-\arccos \frac{4}{5}\right)=\frac{1}{10}(4 \sqrt{3}+3)$.