In a triangle ABC with fixed base BC, the vertex  A moves such that $\cos B+\cos C=4{\sin }^2\frac{A}{2}.$  If a, b, and c denote the lengths of the sides of the triangle opposite to the angles A, B, and C, respectively, then 

Given   $\cos B+\cos C=4{\sin }^2\frac{A}{2}$
or  $\cos \left(\frac{B-C}{2}\right)=2\sin \left(\frac{A}{2}\right)$
or  $2\sin \frac{B+C}{2}\cos \frac{B-C}{2}=4\sin \frac{A}{2}\cos \frac{A}{2}$
or  $\sin B+\sin C=2\sin A$
or  $b+c=2a$
 Thus sum of two variable sides is constant. Hence, the locus of vertex A is an ellipse with B and C as foci