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求解高中数学题,在线等!

函数y=3sin(kx+π/3)的最小正周期t,满足t∈(1,3),求正整数k,并就最小的k值求出其单调区间及对称中心(其中/是分数线,求解题过程和思路!)
被浏览: 0次 2023年02月09日 05:46
热门回答(2个)
游客1

T=2π/k
1<2π/k<3
2π/3K最小=3则y=3sin(3x+π/3)
增区间2kπ-π/2<=3X+π/3<=2Kπ+π/2,即【2Kπ/3-5π/18,2Kπ/3+π/18】
减区间2kπ+π/2<=3X+π/3<=2Kπ+3π/2,

游客2

解;1<2π/k<3, 于是得 2π/3当k=3时, y=3sin(3x+π/3)
单增区间:2kπ/3-5π/18≤x≤2kπ/3+π/18,
单减区间 2kπ/3+π/18≤x≤2kπ/3+7π/18