函数y=2sin(π/6-2x),x∈【0,π】为增函数的区间是？

显示全部楼层 · 2013-10-26 22:53:17

解:1. ∵ y=2sin(π/6-2x)=-2sin(2x-π/6)∴函数y=2sin(π/6-2x)的递增区间即为函数y=2sin(2x-π/6)的递减区间∵x∈[0,π]
∴2x-π/6∈[-π/6,11π/6]令2x-π/6∈[π/2, 3π/2],解得x∈[π/3, 5π/6]即函数y=2sin(π/6-2x)的递增区间为[π/3, 5π/6].2. f(x)=sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)∵x∈[0,π]
∴x+π/4∈[π/4,5π/4]则当x+π/4∈[π/4,π/2],...

千问 · 2013-10-26 22:53:17

x∈【0,π】π/6-2x∈［-11π/6,π/6］
sinx在［-π3/2，-π/2］上是减函数，所以π/6-2x＝-π3/2x=5π/6π/6-2x=-π/2
x=π/3
增函数的区间{π/3,5π/6]...

千问 · 2013-10-26 22:53:17

1y=2sin(π/6-2x),的增区间2kπ+π/2<=π/6-2x<=2kπ+3π/22kπ-3π/2<2x-π/6<2kπ-π/2kπ-2π/3=<x<=kπ-π/6k=1时π/3=<x<=5π/62f(x)=根号2sin(x+π/4)π/4=<x+π/4<π/20<=x<π/4...

千问 · 2013-10-26 22:53:17

1.【π/6，π/3】2.【0，π/4】...