f(x)
=2√3sin(x-π/4)cos(x-π/4)-sin(2x-π)
=√3sin(2x-π/2)-sin(2x-π)
=-√3cos2x+sin2x
=2((1/2)sin2x - √3/2cos2x)
=2sin(2x-π/3)
单调递增区间
2kπ-π/2 <= 2x-π/3 <= 2kπ+π/2
kπ-π/12<= x <= kπ+5π/12k=0,1,2,.