1+2sin(2x+π/3)≠0,解得2x+π/3≠2kπ+7π/6 且 2x+π/3≠2kπ+11π/6
x≠kπ-π/4 且 x≠kπ+5π/12
f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))
设2x+π/3=θ+π/6
f(x)=(√3-2cos(θ+π/6))/(1+2sin(θ+π/6))
=(√3-√3cosθ+sinθ)/(1+√3sinθ+cosθ)
∵sinθ/(1+cosθ)=(1-cosθ)/sinθ=tan(θ/2)
∴由和比定理得f(x)=tan(θ/2)=tan(x+π/12)
∵x≠kπ-π/4 且 x≠kπ+5π/12
∴f(x)≠-√3/3(问了两次呢…………