已知函数f(x)=2根号3sinXcosX+2cos^2X-1(X属于R)

1、f(x)=2√3sinxcosx+2cos^2x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
最小正周期为：T=2π/2=π
∵x属于[0,π/2]
∴2x+π/6属于[π/6,7π/6]
则在区间[0,π/2]上的最大值为2,最小值-1
2、f(x0)=2sin(2x0+π/6)=6/5
∵x0属于[π/4,π/2]
∴2x0+π/6属于[2π/3,7π/6]
即cos(2x0+π/6)=-4/5
cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6
=-4/5*(√3/2)+3/5*(1/2)
=(3-4√3)/10

1)解析：∵函数f（x）=2cos^2x+2√3*sinx*cosx+3=2cos^2x-1+2√3*sinx*cosx+4
=cos2x+√3sin2x+4=2sin(2x+π/6)+4
T=2π/2=π
∴函数f（x）的最小正周期π
(2)解析：∵函数f(x)=2sin(2x+π/6)+4
∵x∈[0，派/3]
f(0)=2sin(2...

f(x)=2根号3sinXcosX+2cos^2X-1
=根号3sin2X+cos2X
=2cos(2X-π/3)
最小正周期=2π/2= π
0<=X<=π/2
-π/3<=2X-π/3<=2π/3
最大值=2 .最小值=1
f(x0)=6/5
2cos(2X-π/3)=6/5
cos(2X-π/3)=3/5