在锐角三角形ABC中,4sin^2[(B+C)/2]-cos2A=7/2,求角A的大小

4sin^2[(π-A)/2]-1+2sin^2(A)=7/2
4cos^2(A/2)+8sin^2(A/2)cos^2(A/2)=9/2
令cos^2(A/2)=x
4x+8(1-x)x=9/2
8x+16x-16x^2=9
16x^2-24x+9=0
(4x-3)^2=0
x=3/4
cos(A/2)=根号3/2
A/2=π/6
A=π/3已知cos(a+p)=4/5,cos(a-p)=-4/5，a+p€(7派/4,2派),a-p€(3派/4)，求cos2a