求函数y=2sin(2x+π/6)(x属于[－π，0])的单调递减区间是

显示全部楼层 · 2013-10-12 11:39:28

y=sin(2x π/3)把2x π/3 看做一个数那么y的单调递减区间为
π/22kπ≤2x π/3≤3π/2 2kπ
k∈z
π/12kπ≤x≤7π/12 kπ
k∈z 与x∈[0,π] 求交集就是所求区间
也就是k=0时
π/12 ≤x≤π...

千问 · 2013-10-12 11:39:28

你画一下正弦函数的图就出来了撒，（2X+π/6)在【-3/2π，-1/2π】区间内该函数单调递减，剩下来你可以求出X的区间了吧？...

千问 · 2013-10-12 11:39:28

y=sint的递减区间为[-3π/2,-π/2]。所以-3π/2<=2x+π/6<=-π/2，解得-π5/6<=x<=-π/3,又因为x属于[－π，0]，所以y=2sin(2x+π/6)的单调递减区间为[-5π/6,-π/3]...