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=...