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.

题目

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、

答案

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】
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