求函数y=3sin(2x+π⼀4）求单调递减区间。要详解答案 我加分

sinx的减区间是(2kπ+π/2,2kπ+3π/2)
所以这里2kπ+π/2<2x+π/4<2kπ+3π/2
2kπ+3π/4<2x<2kπ+7π/4
kπ+3π/8所以减区间是(kπ+3π/8,kπ+7π/8)

y=3sin(2x+π/4)
令z=2x+π/4
sinz函数的单调增区间为2kπ-π/2≤z≤2kπ+π/2
2kπ-π/2≤2x+π/4≤2kπ+π/2
2kπ-3π/4≤2x≤2kπ+π/4
kπ-3π/8≤x≤kπ+π/8
所以y=3sin(2x+π/4)的单调增区间为[kπ-3π/8,kπ+π/8] k∈整数
sinz函数的单调减区间为2kπ+π/2≤z≤2kπ+3π/2
2kπ+π/2≤2x+π/4≤2kπ+3π/2
2kπ+π/4≤2x≤2kπ+5π/4
kπ+π/8≤x≤kπ+5π/8
所以y=3sin(2x+π/4)的单调减区间为[kπ+π/8,kπ+5π/8] k∈整数

答案：（π/8+kπ，5π/8+kπ）（闭区间也可以的）
对于函数y=sinx的单调递减区间是（π/2+2kπ， 3π/2+2kπ),令π/2+2kπ<2x+π/4<3π/2+2kπ,解得π/8+kπ

sinx的减区间是(2kπ+π/2,2kπ+3π/2)
2kπ+π/2<2x+π/4<2kπ+3π/2
2kπ+π/4<2x<2kπ+5π/4
kπ+π/8减区间是(kπ+π/8,kπ+5π/8)