求函数y=2sin(x﹣π/3)的单调区间

Question

求函数y=2sin(x﹣π/3)的单调区间

Answer 1

单调增区间：-π/2+2*k*pi<=x-π/3<=π/2+2*k*π,解出x的范围,再用区间表示即可.
单调减区间：π/2+2*k*pi<=x-π/3<=(3*π)/2+2*k*π,解出x的范围,再用区间表示即可.