Question

函数y=sin(2x-π÷3)-sin2x的一个单调递增区间详解过程

Answer 1

解：y=sin(2x-π/3)-sin2x
=sin2xcos2∏/3-cos2xsin∏/3-sin2x
=1/2sin2x-cos2xsin∏/3-sin2x
=-(sin2xcos2∏/3+cos2xsin∏/3)
=-sin（2X+π/3）
所以函数y=sin(2x-π÷3)-sin2x的一个单调递增区间为y=-sin(2x+∏/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}

追问

谢谢了

追答

2Kπ+π/2<=（2X+π/3）<=2Kπ +3π/2
2kπ+π/6<=2x<=2kπ+7π
Kπ+π/12<=x<=Kπ+7π/12

Answer 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

追问

谢谢了

追答

单调递增区间:
2kπ<=2x-π/6<=2kπ+π
2kπ+π/6<=2x<=2kπ+π+π/6
2kπ+π/6<=2x<=2kπ+7π/6
kπ+π/12<=x<=kπ+7π/12

Answer 3

解：y=sin2x-sin(2x-π/3) =2cos(π/3)sin2x-[sin2xcos(π/3)-则不妨令k=0,可得函数的一个单调递增区间为[-5π/12, π/12] 1