Question

已知f(x)=5cos²x+sin²x-4√3sinxcosx

1.化简f(x)的解析式，并求出f(x)的最小正周期
2.当x属于[-π/6,π/4]时，求f(x)的值域

Answer 1

1,f(x)=5(cosx)^2+(sinx)^2-4√3sinxcosx
=4(cosx)^2-2√3sin2x+1
=2cos2x-2√3sin2x+3
=4cos(2x+π/3)+3
最小正周期为T=2π/2=π。
2,-π/6<=x<=π/4，则0<=2x+π/3<=5π/6。
f(x)最小值是4cos5π/6+3=3-2√3，最大值是4cos0+3=7。
f(x)的值域是[3-2√3,7]。

追问

则0<=2x+π/3<=5π/6怎么来的

追答

-π/6<=x<=π/4，-π/3<=2x<=π/2，-π/3+π/3<=2x+π/3<=π/2+π/3，0<=2x+π/3<=5π/6。