已知函数f(x)=2cosxcos(x-兀/6)-亅

Question

已知函数f(x)=2cosxcos(x-π/6)-根号3乘sinxcosx1)求f(x)的最小正周期2）当x属于[0,π]时,若f(x)=1,求x的值π为圆周率派

Answer 1

是这样的吧?
f(x)=2cosxcos(x-π/6)-根号3sin^2x+sinxcosx
=2cosxcos(x-π/6)-根号3sin^2x+sinxcosx
=2cosx(根号3/2cosx+1/2sinx)-根号3in^2x+sinxcosx
=根号3cos^2x+sinxcosx-根号3sin^2x+sinxcosx
=根号3(cos^2x-sin^2x)+2sinxcosx
=根号3cos2x+sin2x
=2（根号3/2cos2x+1/2sin2x)
=2sin(2x+π/3)
最小正周期为T=2π/2=π
（2）2sin(2x+π/3)=1
又2X+π/3属于[π/3,7π/3],所以,2X+π/3=5π/6或7π/6
那么,X=π/4 或5π/12