已知函数f（x）=cos（2x+π/3）+sin2x-cos2x+2√3sinxcosx

f（x）=cos（2x+π/3）+sin2x-cos2x+2√3sinxcosx
=cos2xcosπ/3-sin2xsinπ/3+sin2x-cos2x+√3sin2x
=sin2x-(cos2x-√3sin2x)/2
=sin2x-sin(π/6-2x)
=2cos[(2x+π/6-2x)/2]*sin[(2x-π/6+2x)/2]
=2cos(π/12)sin(2x-π/12)
单增 2kπ-π/2 ≤2x-π/12≤2kπ+π/2
单减 2kπ+π/2 ≤2x-π/12≤2kπ+3π/2
其它的自己做

1f'(x)=-2sin(2x+π/3)+2cos2x+2sin2x+2√3cos2x=2sin(2x+π/3)+2cos2x
1令f'(x)<0，得到sin(2x+π/3)+cos2x<0,由sin α-sin β=2cos[(α+β)/2]·sin[(α-β)/2] 得到cos(2x-5π/12)<0,因此，π/2+2kπ<2x-5π/12<3π/2+2kπ，因此减区间是：（11π/...