已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos（B+C）】-cos2A=7/2

先解出cosA
因为cos(B+C) = cos(π -A) = - cosA
cos2A = 2cos²A -1
解方程
2(1+cosA) -(2cos²A-1) = 7/2
化简即　2cos²A - 2cosA + 1/2 =0
cosA = 1/2
(1)
2sinBcosC = sin(B+C) - sin(C-B) = sinA - sin(C-B)
所以　 sinA - sin(C-B)= sinA
sin(C-B) =0
C=B
又因为A=π/3 , 所以C=B= (π-π/3)/2=π/3
ABC是等边三角形
(2)
余弦定理\
a²= b² +c² -2bc cosA = b²+c² -bc
所以　3 = b²+c² -bc
b+c=3, 所以(b+c)² = b²+c² +2bc =9　
两式相减
3bc = 9-3 =6
bc =2
故bc =2,
b+c =3
解之得
b=1, c=2或
b=2, c=1