f(x)=2sin^2(x+π/4)-cos(2x+π/6)=2*[(1-cos(2x+π/2))/2]-cos(2x+π/6)=1-cos(2x+π/2)-[cos2xcos(π/6)-sin(2x)sin(π/6)]=1+sin2x-(√3/2)cos2x+(1/2)sin2x=(3/2)sin2x-(√3/2)cos2x+1=√3sin(2x-π...