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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.
人气:436 ℃ 时间:2019-08-21 05:49:25
解答
(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}
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