Question

已知函数f(x)=(sinx-cosx)sin2x/sinx，求f(x)单调减区间

Answer 1

解：（1）由sinx≠0得x≠kπ（k∈Z），
故求f（x）的定义域为{x|x≠kπ，k∈Z}．
∵f（x）=(sinx-cosx)sin2x /sinx
=2cosx（sinx-cosx）
=sin2x-cos2x-1
=√2sin（2x-π/4 )-1
函数y=sinx的单调递减区间为[2kπ+π /2 ，2kπ+3π/2 ]（k∈Z）
∴由2kπ+π /2 ≤2x-π/4≤ 2kπ+3π/2 ,x≠kπ（k∈Z）
得kπ+π 3/8≤x≤kπ+7π/8 (k∈Z）
∴f（x）的单调递减区间为：[kπ +3π /8,kπ+7π/8】（ k∈Z）

Answer 2

f(x)=(sinx-cosx)sin2x/sinx
=2(sinx-cosx)sinxcosx/sinx
=2(sinx-cosx)cosx
=2sinxcosx-2cos²x
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
π/2+2kπ<2x-π/4<3π/2+2kπ
3π/4+2kπ<2x<7π/4+2kπ
3π/8+kπ<x<7π/8+kπ
所以，f(x)的递减区间为（3π/8+kπ，7π/8+kπ）k∈Z

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Answer 3

f(x)=(sinx-cosx)sin(2x)/sinx
= 2(sinx-cosx)cosx
= sin2x-(cos2x+1)
=√2( sin(2x-π/4) -1
单调减区间
2kπ+π/2<=2x-π/4<= 2kπ+3π/2
kπ+3π/8<=x<=kπ+7π/4 ; k=0,1,2,...

Answer 4

解决方案：
（1）sinx的≠0，X≠Kπ，K∈Z

F（x）的=（sinx的cosx的）sin2x/sinx

=（sinx的cosx的）* 2cosx BR /> = 2sinxcosx-2cos 2所述

= sin2x - cos2x-1

=√2sin（2X-π/ 4）下-1

T =2π/ 2 =π

（2）增加范围≤2X-PI

2kπ-π/ 2/4 <2kπ或2kπ<2X-π/ 4≤2kπ+ PI / 2

2kπ-π/ 4≤2倍<2kπ+π/ 4或2kπ+π/ 4 <2倍≤2kπ

Kπ-π/ 8≤X <Kπ+ PI / 8或Kπ+ PI / 8 3/4的PI <X≤ Kπ+3 PI / 8
增加间隔[Kπ-π/ 8Kπ+π/ 8）和（Kπ+ PI / 8，Kπ的+3 PI / 8}，K∈