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(C) In $	riangle ABC$,we have $A + B + C = 180^{\circ}$,so $\frac{B+C}{2} = 90^{\circ} - \frac{A}{2}$.Given $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$.

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$(A, C)$

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(C) In $	riangle ABC$,we have $A + B + C = 180^{\circ}$,so $\frac{B+C}{2} = 90^{\circ} - \frac{A}{2}$.Given $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$.Using the sum-to-product formula: $2 \cos \frac{B+C}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$.Since $\cos \frac{B+C}{2} = \sin \frac{A}{2}$,we have $2 \sin \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$.Dividing by $2 \sin \frac{A}{2}$ (assuming $A 
eq 0$),we get $\cos \frac{B-C}{2} = 2 \sin \frac{A}{2}$.Multiplying both sides by $2 \cos \frac{A}{2}$,we get $2 \cos \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin \frac{A}{2} \cos \frac{A}{2}$.Using $2 \cos \frac{A}{2} = 2 \sin \frac{B+C}{2}$,we get $2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} = 2 \sin A$.This simplifies to $\sin B + \sin C = 2 \sin A$.By the Sine Rule,$\frac{b}{2R} + \frac{c}{2R} = 2 \frac{a}{2R}$,which implies $b + c = 2a$.Since $b + c = AB + AC = 2BC$,and $BC$ is a fixed base,the sum of the distances of $A$ from two fixed points $B$ and $C$ is constant $(2BC > BC)$.Therefore,the locus of point $A$ is an ellipse with foci at $B$ and $C$.

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