Question

已知函数f(x)=4cos·sin(x+(π/6)) (1)求f(x)的最小正周期及最大值 （2）求f(x)的单调递增区

Answer 1

f(x)=4cosx*sin（x+π/6)

=4cosx(sinx*√3/2+cosx*1/2)

=2√3sinxcosx+2(cosx)^2

=√3sin2x+cos2x+1

=2sin(2x+π/6)+1

f(x)的最小正周期 T =π

最大值为2+1=3

2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z.

kπ-π/3≤x≤kπ+π/6,k∈Z.

f(x)的单调递增区间是[kπ-π/3,kπ+π/6],k∈Z.