1由余弦定理,
cosx=（a^2+c^2-b^2)/2ac
=(a^2+c^2)/2ac-1/2
>=1-1/2=1/2
从而x的取值范围为（0,Pi/3]
2
f(x)=3^0.5*1/2*sin2wx-(1+cos2wx)/2
=[3^0.5/2*sin2wx-1/2cos2wx]-1/2
=sin(2wx-Pi/6)-1/2
由周期为pi/2知2w=4
4x-Pi/6的取值区间为（-Pi/6,7Pi/6]
从而sin(2wx-Pi/6)取值区间为[-1/2,1]
f(x)的值域为[-1,1/2]