f(x)=3sin(2x+π/2)
令2kπ+π/2＜2x+π/2＜2kπ+3π/2(k∈Z)
得kπ＜x＜kπ+π/2(k∈Z)
即函数f(x)=3sin(2x+π/2)的单调递减区间是(kπ,kπ+π/2)(k∈Z)
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