Question

函数y＝sin(π/3-1/2χ)，π∈［0，4π］，求函数的单调递减区间

Answer 1

原题是:函数y＝sin(π/3-x/2),x∈[0,4π],求函数的单调递减区间.

y=sin(π/3-x/2)=-sin(x/2-π/3)
=sin(x/2-π/3+π)=sin(x/2+2π/3)
即y=sin(x/2+2π/3)
x∈[0,4π],x/2+2π/3∈[2π/3,2π+2π/3]
当2π/3≤x/2+2π/3≤2π/2或2π+π/2≤x/2+2π/3≤2π+2π/3
即0≤x/≤5π/3或11π/3≤x≤4π时,x在单减区间中.
所以原函数的单减区间是[0,5π/3],[11π/3,4π]

希望能帮到你！

Answer 2

我做做，等等

追答

不好意思噶，答案和下边的不一样，一个区间都没做出来😂😂😂😂

追问

没事儿

Answer 3

(-1/3π-4kπ,3/5π-4kπ) (k∈Z)