Question

已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos²x(x∈R)。1,求函数F(x)的最大值

2,若f(a+b)=1,求a+b的值
3求使f（x）大于等于二的x的取值范围
注：求详细过程

Answer 1

f(x)=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+2cos²x
=2sin2xcosπ/6+1+cos2x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
1.f(x)最大值为2+1=3
2.f(a+b)=2sin[2(a+b)+π/6]+1=1
sin[2(a+b)+π/6]=0
2(a+b)+π/6=kπ
a+b=kπ/2-π/12 k=0.1.2.....
3.f(x)≥2
f(x)-2=2sin(2x+π/6)-1≥0
sin(2x+π/6)≥1/2
x∈[kπ,π/3+kπ]，k=0.1.2....

Answer 2

f(x)=sin2xcosπ/6+sinπ/6cos2x+sin2xcosπ/6-sinπ/6cos2x+2cos²x
=2cosπ/6sin2x+cos2x+1
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
1、f(x)最大值=2+1=3；
2、f(a+b)=2sin(2(a+b)+π/6)+1=1;
2(a+b)+π/6=0+kπ(k∈Z)
a+b=-π/12+kπ/2(k∈Z)
3、f(x)=2sin(2x+π/6)+1≥2;
sin(2x+π/6)≥1/2;
π/6+2kπ≤2x+π/6≤5π/6+2kπ;
0+2kπ≤2x≤4π/6+2kπ;
0+kπ≤x≤2π/6+kπ;(k∈Z)