函数f(x)=1+2sin(2x-π/3)单调区间

题目

函数f(x)=1+2sin(2x-π/3)单调区间

答案

解当2kπ-π/2≤2x-π/3≤2kπ+π/2,k属于Z时,f(x)=1+2sin(2x-π/3)增函数,
即当kπ-1π/12≤x≤kπ+5π/12,k属于Z时,f(x)=1+2sin(2x-π/3)增函数
即f(x)=1+2sin(2x-π/3)的增区间是[kπ-1π/12,2kπ+5π/12],k属于Z时.
当2kπ+π/2≤2x-π/3≤2kπ+3π/2,k属于Z时,f(x)=1+2sin(2x-π/3)减函数,
即当kπ+5π/12≤x≤kπ+11π/12,k属于Z时,f(x)=1+2sin(2x-π/3)减函数
即f(x)=1+2sin(2x-π/3)的减区间是[kπ+5π/12,2kπ+11π/12],k属于Z时.