6. In $\triangle A B C$, $\frac{\cos A+2 \cos C}{\cos A+2 \cos B}=\frac{\sin B}{\sin C}$, determine the shape of $\triangle A B C$.
6. From the known, we have
$$
\cos A \sin \frac{B-C}{2}\left(\cos \frac{B+C}{2}-2 \cos \frac{B-C}{2}\right)=0 \text {. }
$$

Since $0 < \frac{|B-C|}{2}$, we have $\cos \frac{B+C}{2} < \cos \frac{B-C}{2} < 2 \cos \frac{B-C}{2}$, which means $\cos \frac{B+C}{2} - 2 \cos \frac{B-C}{2} \neq 0$. Therefore, we deduce that $\cos A = 0$ or $\sin \frac{B-C}{2} = 0$. Hence, $A = 90^{\circ}$ or $B = C$, which means $\triangle ABC$ is a right triangle or an isosceles triangle.