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,則cos(θ+π/6)=1/2,
因為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,
函數(shù)的遞減區(qū)間是[2kπ-π/6,2kπ+5π/6] ,k∈Z.
2kπ-π≤x+π/6≤2kπ,k∈Z.
2kπ-7π/6≤x≤2kπ-π/6,
函數(shù)的遞增區(qū)間是[2kπ-7π/6,2kπ-π/6] ,k∈Z.