f(x)=sin(2x+π/6)-cos2x+1
所以为2π/2=π
f(x)=根号3/2 sin2x-(cos2x)/2+1
=sin(2x-π/6)+1
所以最大值为2,x=π/2+2kπ-π/6=π/3+2kπ
由上式,[-π/2+2kπ,π/2+2kπ]