求函数y=sin2x-根号3·cos2x的单调递增区间
问题描述:
求函数y=sin2x-根号3·cos2x的单调递增区间
答
y=sin2x-根号3·cos2x
= 2(sin2x cosπ/3 - cos2x sinπ/3)
= 2sin(2x-π/3)
当2x-π/3∈【2kπ,2kπ+π),即x∈【kπ+π/6,kπ+2π/3),其中k∈Z时,单调递增为什么2x-π/3∈【2kπ,2kπ+π)?而不是2x-π/3∈【2kπ-π/2,2kπ+π/2】然后是πk-π/12