Question

三角函数的三条题目.（急）1、若f(x)=2sinωx(0

Answer 1

1.先求一下单调区间

令2kπ-π/2≤ωx≤2kπ+π/2

解之得到单调递增区间为[2kπ/ω-π/2ω,2kπ/ω+π/2ω]

令k=0,得到原点附近的单调性：f(x)在[-π/2ω,π/2ω]上单调递增

因0π/3

故f(x)在[0,π/3]上单调递增

故f(π/3)=√2

即2sinωπ/3=√2

故ω=6k+3/4或6k+9/4

结合0