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在(π/2+2kπ，3π/2+2kπ)之间是单调递减的！所以把2x+π/4看做一个整体可以知道π/2+2kπ<(2x+π/4)<3π/2+2kπ解出单调递减区间为π/8+kπ评论00加载更多
