f(x)=2sin^2(pi/4+x)-√3cos2x
=[1-cos(2(pi/4+x)]-根号3 cos2x
=1-cos(pi/2+2x)-根号3cos2x
=1+sin2x-根号3cos2x
=1+2[1/2sin2x-根号3 /2 cos2x]
=1+2sin(2x-60)