题目

f(x)=2sin(x+ θ/2).cos(x+ θ/2)+2倍根号三cos²(x+ θ/2)-根号三
（1）化简f（x）；（2）若 θ∈[0,π],求使f（x）成为偶函数的 θ取值；（3）在（2）成立的条件下,求满足f（x）=1,且x∈[-π,π]的x的集合.

提问时间：2020-10-30

答案

(1) f(x)=2sin(x+ θ/2).cos(x+ θ/2)+2倍根号三cos²(x+ θ/2)-根号三
=sin(2x+ θ)+√3[cos(2x+ θ)+1]-√3
=sin(2x+ θ)+√3[cos(2x+ θ)=2sin(2x+ θ+π/3)
(2) 若 θ∈[0,π],使f（x）成为偶函数
θ+π/3=π/2,θ=π/6
(3) f(x)=2sin(2x+ θ+π/3)=2sin(2x+π/6 +π/3)
=2cos2x=1
cos2x=1/2,2x=2kπ±π/3,
x=kπ±π/6,且,x∈[-π,π],
∴x=±π/6,x=±5π/6
x∈{-π/6,π/6,-5π/6,5π/6}