Question

y=2sin(1/2x+π/3)单调区间求大神解出来

RT

Answer 1

2kπ-π/2<1/2x+π/3<2kπ+π/2
4kπ-5π/3<x<4kπ+π/3
因此，单增区间：[4kπ-5π/3, 4kπ+π/3]
单减区间：[4kπ+π/3, 4kπ+7π/3]

Answer 2

y=2sinx的单调递增区间是[-π/2+2kπ,π/2+2kπ](k∈Z)
由复合函数的同增异减性质
则由 -π/2+2kπ ≤1/2 x+π/3≤π/2+2kπ
解得-π5/3 +4kπ≤x≤π/3 +4kπ
所以y=2sin（1/2 x+π/3）的单调递增区间是[-5π/3+4kπ,π/3+4kπ](k∈Z)
y=2sinx的单调递减区间是[π/2+2kπ,3π/2+2kπ](k∈Z)
由 π/2+2kπ ≤1/2 x+π/3≤3π/2+2kπ
解得π/3 +4kπ≤x≤7π/3 +4kπ
所以y=2sin（1/2 x+π/3）的单调递减区间是[π/3+4kπ,7π/3+4kπ](k∈Z)