已知函数f(x)=2cosx/2 (√3cosx/2-sinx/2)；⑴设θ∈[-π/2,π/2],且f(θ)=√3+1 求θ的值 ⑵求单调区间

题目

已知函数f(x)=2cosx/2 (√3cosx/2-sinx/2)；⑴设θ∈[-π/2,π/2],且f(θ)=√3+1 求θ的值 ⑵求单调区间
已知函数f(x)=2cosx/2 (√3cosx/2-sinx/2)；1）设θ∈[-π/2,π/2],且f(θ)=√3+1 求θ的值 2）求f(x)的单调区间

答案

f(x)=2cosx/2 (√3cosx/2-sinx/2)
=2√3cos²(x/2)-2 sinx/2 cosx/2
=√3(1+cosx)-sinx
=√3 cosx-sinx+√3
=2 cos(x+π/6) +√3
(1)f(θ)=√3+1,则cos(θ+π/6)=1/2,
因为θ∈[-π/2,π/2],
所以θ+π/6=π/3或-π/3
θ=π/6或-π/2.
(2)
2kπ≤x+π/6≤2kπ+π,k∈Z.
2kπ-π/6≤x≤2kπ+5π/6,
函数的递减区间是[2kπ-π/6,2kπ+5π/6] ,k∈Z.
2kπ-π≤x+π/6≤2kπ,k∈Z.
2kπ-7π/6≤x≤2kπ-π/6,
函数的递增区间是[2kπ-7π/6,2kπ-π/6] ,k∈Z.