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/3θ...