若函数f（x）=2√3sinxcosx+2cos²x+m在区间[0,兀½]上的最大值为6,求常数m的取值及此函数当x∈R时的最小值,并求对应的x取值的集合

f（x）=2√3sinxcosx+2cos²x+m=√3sin2x+cos2x+1+m=2sin(2x+π/6)+1+m在区间x∈[0,兀½] 2x+π/6∈[π/6,7π/6] sin(2x+π/6)最大值=1所以 2+1+m=6m=3f(x)=2sin(2x+π/6)+42x+π/6=7π/6 sin(2x+π/6)最小...