Question

求一道函数体: (1)求函数y=3cos(2x-π/3),x∈R的单调区间; (2)求函数y=sin(-3x+π/4),x∈R的单调区间.

求详细过程~
谢谢啦、

Answer 1

1. y=cosx 单调减区间为 [2kπ,2kπ+π]
函数y=3cos(2x-π/3),x∈R的单调区间
2kπ<=2x-π/3<=2kπ+π 解得
单调减区间 【kπ+π/6，kπ+2π/3】

y=cosx 单调增减区间为 [2kπ-π,2kπ]
函数y=3cos(2x-π/3),x∈R的单调增区间
2kπ-π<=2x-π/3<=2kπ 解得
单调增区间 【kπ-2π/3，kπ+π/6】

2. y=sinx 单调增区间为 [2kπ-π/2,2kπ+π/2]
函数y=sin(-3x+π/4)=-sin(3x-π/4)
函数y=sin(-3x+π/4)=-sin(3x-π/4) 单调减区间为
2kπ-π/2<=3x-π/4<=2kπ+π/2
单调减区间为 【2kπ/3-π/12，2kπ/3+3π/12】

y=sinx 单调减区间为 [2kπ+π/2,2kπ+3π/2]
函数y=sin(-3x+π/4)=-sin(3x-π/4) 单调增区间为
2kπ+π/2<=3x-π/4<=2kπ+3π/2
单调增区间为 【2kπ/3+3π/12，2kπ/3+7π/12】

πππ

Answer 2

(1)递增区间：令 -π+2kπ <=2x-π/3<= 0+2kπ k=0,1,2,3.......
递减区间：令 0+2kπ <=2x-π/3<= π+2kπ k=0,1,2,3.......

（2）递增区间：令 -π/2+2kπ <=-3x+π/4<= π/2+2kπ k=0,1,2,3.......
递减区间：令 π/2+2kπ <=(-3x+π/4<= 3π/2+2kπ k=0,1,2,3.......
求出x的范围即可