已知函数f x=3cos(2X+π/4),求单调递减区间

f(x)=3cos(2x+π/4)
f'(x)=-6sin(2x+π/4)
令：f(x)＜0,即：-6sin(2x+π/4)＜0
整理,得：sin(2x+π/4)＞0
解得：kπ-π/8＜x＜kπ+3π/8,k=0、±1、±2、±3、……
即：函数的单调递减区间是：x∈(kπ-π/8,kπ+3π/8),k=0、±1、±2、±3、……

-π/8+kπ，π3/8+kπ

cosx的递减区间是 [2kpi,pi+2kpi]
2kpi<=2x+pi/4<=pi+2kpi
-pi/8+kpi<=x<=3pi/8+kpi
[-pi/8+kpi,3pi/8+kpi]