fx=3sin(2x-π/3)-1
当 sin(2x-π/3)=1时 取得最大值为3-1=2
这时 2x-π/3=2kπ+π/2
得 x=kπ+5π/12 k∈z求此时函数的递减区间单调减区间2x-π/3∈[2kπ+π/2,2kπ+3π/2]x∈[kπ+5π/12,2kπ+11π/12]所以单调减区间为 [kπ+5π/12,2kπ+11π/12] k∈z