y=sin(2x-π⼀3)-sin2x的一个单调递增区间是

y=3sin(π/6-3x)=-3sin(3x-π/6),x∈[-π/2,π/2]的单调递增区间即y=3sin(3x-π/6)x∈[-π/2,π/2]的单调递减区间,由2kπ π/2<=3x-π/6<=2kπ 3π/2,2kπ/3 2π/9<=x<=2kπ/3 5π/9,
k=-1,0时-4π/9<=x<=-π/9，2π/9<=x<=5π/9，结合x∈[-π/2,π/2]得
y=3sin(π/6-3x)x∈[-π/2,π/2]的单调递增区间为：[-4π/9，-π/9]，[2π/9，π/2]

y=sin(2x-π/3)-sin2x
=2 cos[(2x-π/3+2x)/2] sin[(2x-π/3-2x)/2]
=2 cos(2x-π/6) sin(-π/6)
=- cos(2x-π/6)
=- cos(2x-π/6)
单调递增区间:
2kπ<=2x-π/6<=2kπ+3π/2
2kπ+π/6<=2x<=2kπ+3π/2+π/6
2kπ+π/6<=2x<=2kπ+5π/3
kπ+π/12<=x<=kπ+5π/6

展开并化简得到y=-sin（2X+π/3）
求y=sin(2x-π/3)-sin2x的一个单调递增区间即求y=-sin（2X+π/3）的减区间
当2Kπ+π/2<=（2X+π/3）<=2Kπ
+3π/2时
y=sin(2x-π/3)-sin2x的一个单调递增区间为{Kπ+π/12，Kπ+7π/12}