Question

已知函数f(x)=根号3sin05x+sinxcosx,x∈[π/2,π]

1 求f（x）的零点 2 求f（x）的最大值和最小值

Answer 1

f(x)=√3*(1-cos2x)/2+1/2*sin2x
=sin2xcosπ/3-cos2xsinπ/3+√3/2
=sin(2x-π/3)+√3/2
令f(x)=0
sin(2x-π/3)=-√3/2
2x-π/3=2kπ-π/3,2x-π/3=2kπ-2π/3
所以零点是x=kπ/3,x=kπ/3-π/6

f(x)=sin(2x-π/3)+√3/2
所以最大值是1+√3/2
最小值是-1+√3/2