Question

F(x)=(sinx一cosx)sin2x/sinx 单调减区间

再次等数学高手

Answer 1

解：
F(x)=(sinx一cosx)sin2x/sinx=2(sinx一cosx)cosx=sin2x-cos2x-1=√2sin(2x-π/4)
所以F(x)单调减区间为：π/2+2kπ<=2x-π/4<=3π/2+2kπ，即3π/8+kπ<=x<=7π/8+kπ,(k∈Z)
或者表示为[3π/8+kπ,7π/8+kπ](k∈Z)

Answer 2

注意到，原函数的定义域为：x≠kπ（k=0,1,2…）
F(x)=(sinx一cosx)sin2x/sinx
=2cosx(sinx一cosx)
=2sinxcosx-2cos²x
=sin2x-cos2x-1
=√2sin(2x+3π/4)-1
求导：F'(x)=2√2cos(2x+3π/4)
满足F'(x)<0，解得x即为原函数的单调减区间：
易知，cos(2x+3π/4)的单调减区间为kπ≤2x+3π/4≤(k+1)π
(kπ-3π/4)/2≤x≤(k+π/4)/2
解得F(x)的单调区间为：[(kπ-3π/4)/2,k+π/4)/2]

Answer 3

sinx≠0,x≠kπ;
F(x)=2cosx(sinx-cosx)=sin2x-(1+cos2x)
=-1-cos2x+sin2x=-1+√2sin(2x-π/4)
π/2+2kπ≤2x-π/4≤π3/2+2kπ
3π/8+kπ≤x≤π7/8+kπ
所以F(x)的减区间为[3π/8+kπ,π7/8+kπ]

Answer 4

F(x)=(sinx-cosx)*2sinxcosx/sinx
=(sinx-cosx)*2cosx
= 2sinxcosx-（1+cos2x）
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
-π+2Kπ <2x-π/4<-π/2+2Kπ ①
π/2+2Kπ<2x-π/4<π+2kπ ②
不等式组①②自己解X吧，打字太慢了