Question

已知函数f(x)=sinx^2+√3sinxcosx+2cosx^2,x属于R

1.求函数f（x）的最小正周期和单调增区间
2.函数f（x）的图像可以由函数y=sin2x(x属于R)的图像经过怎样的变换得到？

要过程~~谢谢~~

Answer 1

f(x)=1+√3sin2x/2+cos2x/2+1/2=3/2+sin(2x+π/6);
T=2π/2=π;
-π/2+2kπ≤2x+π/6≤π/2+2kπ;(k∈Z)
-2π/3+2kπ≤2x≤π/3+2kπ;
-π/3+kπ≤x≤π/6+kπ;(k∈Z)
单调增区间为[-π/3+kπ,π/6+kπ](k∈Z);
2、f（x）=3/2+sin(2(x+π/12));
∴往左平移π/12个单位即可

Answer 2

f(x)=(1-cos2x)/2+√3/2sin2x+cos2x+1
=1/2cos2x+√3/2sin2x+3/2
=sin(2x+π/6)+3/2
1. 最小正周期T=2π/2=π
单调增区间

2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
单调增区间 [kπ-π/3，kπ+π/6] k∈Z
2. 函数y=sin2x向左平移π/12个单位得到函数 y=sin(2x+π/6)
再向上平移3/2个单位得到函数 y=sin(2x+π/6)+3/2