y=2sinwx (w>0)先求出函数在原点附近的递增区间由-π/2≤wx≤π/2的-π/(2w)≤x≤π/(2w)即y=sinwx在原点附近的递增区间为[-π/(2w),π/(2w)]因为函数在[-π/3,π/4]上单调递,所以[-π/3,π/4]是[-π/(2w),π/(2w)]...