求y=2sin(-2x-π⼀6)的单调增区间

2025-12-15 02:58:00
推荐回答(1个)
回答1:

y=2sin(-2x-π/6)
=-2sin(2x+π/6)
当π/2 +2kπ<2x+π/6<3/2 π+2kπ时,
即π/6 +kπ<x<2π/3 +kπ时单调递增,所以单调增区间是(π/6 + kπ, 2π/3 + kπ)