Question

函数y=sin(-3x+π/4),x∈R,在什么区间上是增函数

为什么我算后是[π/4+2/3kπ，7/12π+2/3kπ]，可答案是[-π/12+2/3kπ，π/4+2/3kπ],

Answer 1

因为函数y=sinx在[-π/2+2kπ，π/2+2kπ]，所以-π/2+2kπ《-3x+π/4《π/2+2kπ，得-3π/4+2kπ《-3x《π/4+2kπ,解得-π/12-2/3kπ《x《π/4-2/3kπ,因为k属于整数，所以π/12+2/3kπ《x《π/4+2/3kπ

追问

是-π/12啊。。。。

追答

因为函数y=sinx在[-π/2+2kπ，π/2+2kπ]，所以-π/2+2kπ《-3x+π/4《π/2+2kπ，得-3π/4+2kπ《-3x《π/4+2kπ,解得-π/12-2/3kπ《x《π/4-2/3kπ,因为k属于整数，所以-π/12+2/3kπ《x《π/4+2/3kπ 打漏了，对不起