f(x)=4cos(wx-π/6)sinwx-cos(2wx+π),若f(x)在【-3π/2,π/2】上为增函数,求W的

函数f(x)=4cos(wx-π/6)sinwx-cos(2wx+π)=2(sin(2wx-π/6)+sin(π/6))+cos2wx
=2sin(2wx-π/6)+cos2wx+1=2(sin2wxcos(π/6)-cos2wxsin(π/6))+cos2wx+1
=sqrt(3)sin2wx+1
题目中函数f(x)的增区间是[-3π/2,π/2]
说明函数的至少1/4个周期要包含x=-3π/2,即函数的最小正周期最小为6π
此时2π/2w=6π,即w=1/6,此时的w最大,即满足题意的w最大值为1/6
此时,f(x)值域为[1-sqrt(3),(sqrt(3)/2)+1]