f(x)=cos(2x+π/3)+sin²x
=(1/2)cos2x-(√3/2)sin2x+(1-cos2x)/2
=-(√3/2)sin2x+1/2
当sin2x=-1时,f(x)有最大值,f(x)max=(1+√3)/2
此时,2x=-π/2+2kπ
即：x=-π/4+kπ
递增区间：
π/2+2kπ