f(x)=2cos^x+根號(hào)3sin2x=cos2x+1+根號(hào)3sin2x=2(sin2xcosPai/6+sinPai/6cos2x)+1=2sin(2x+Pai/6)+1
那么最小正周期T=2Pai/2=Pai
-Pai/2+2kPai<=2x+Pai/6<=Pai/2+2kPai
-Pai/3+kPai<=x<=Pai/6+kPai
故在[0,Pai]上的增區(qū)間是[0,Pai/6]U[2Pai/3,Pai]
(2)a^2+b^2-c^2>=ab
cosC=(a^2+b^2-c^2)/2ab>=ab/2ab=1/2
故：0f(C)=2sin(2C+Pai/6)+1
Pai/6<=2C+Pai/6<=5Pai/6
故1/2<=sin(2C+Pai/6)<=1
即有：2*1/2+1<=f(C)<=2*1+1
即有：2<=f(C)<=3