Question

急！！！求答案，在线等！！！

Answer 1

f(x)=sin^2(ωx)+√3sin(ωx)cos(ωx)
=[1-cos(2ωx)]/2+(√3/2)sin(2ωx)
=(√3/2)sin(2ωx)-(1/2)cos(2ωx)+(1/2)
=sin(2ωx-π/6)+(1/2)
T=π=2π/(2ω)
ω=1
f(x)=sin(2x-π/6)+(1/2)
由-π/2+2kπ≤2x-π/6≤π/2+2kπ得：
-π/6+kπ≤x≤π/3+kπ
单调增区间：[-π/6+kπ，π/3+kπ]
0≤x≤π
-π/6≤2x-π/6≤11π/6
-1≤sin(2ωx-π/6)≤1
-1/2≤sin(2ωx-π/6)+(1/2)≤3/2
值域为：[-1/2 , 3/2]

Answer 2

会发现G8II好吃发图看嘛v复古人哇沙发一卡卡

追问

？

Answer 3

sin²wx是什么？中？