F(x)=2sinxcosx+2倍根号3cosx的平方+m.(1)当x属于R时求函数f(x)单调递增区间（2）当x大于等

f（x）=2sinxcosx+2√3 cos^2x-√3=2sinxcosx+√3(2cos^2x-1)=sin2x+√3cos2x=2(1/2sin2x+√3/2cos2x)=2sin(2x+π/3)所以f(x)=2sin(2x+π/3)当2x+π/3=-π/2+2kπ(k∈Z),即x=-5π/12+kπ(k∈Z),f(x)最小值=-2 最小正...

f(x)=sin2x+√3cos2x=2sin(2x+π/3),g(x)=mcos(2x-π/6)+3-2m
∵存在x1,x2∈[0,π/4],f(x1)=g(x2)
∴f(x)在x∈[0,π/4]内的最大值大于g(x)的最小值且最小值小于g(x)的最大值
∴f(x)最大=f(π/12)=2≥g(x)最小=g(π/4)=3-3/2m
f(x)最小=f(π/4)=1≤g(x)最大=g(π/12)=3-m
∴2/3≤m≤2

f(x)=2sinxcosx+2√3cos^2x-√3
=sin(2x)+√3(2cos^2x-1)
=sin(2x)+√3cos(2x)
=2[sin(2x)/2+√3cos(2x)/2]
=2sin(2x+60°)

f(x)=2sinxcosx+2√3cos²x-√3
=2sinxcosx+√3(2cos²x-1)
=sin2x+√3cos2x
=2sin(2x+π/3)
最小正周期T=2π/2=π
-π/2+2kπ

f(x)=2sinxcosx+2√3cos^2-√3
=sin2x+√3(2cos^2-1)
=sin2x+√3cos2x
=2sin(2x+60°)
60°