令t=2+cosx,则cosx=t-2,t的取值范围是[1,3]且t≠2
(cosx)^2=(t-2)^2,(sinx)^2=1-(cosx)^2=1-(t-2)^2
带入f'表达式整理得：
f'(t)=-(t-2)^2+1/(t-2)^2=-t^2+4t-4+1/(t-2)^2
所以f(t)=-(1/3)t^3+2t^2-4t-1/(t-2)
即f(x)=-(1/3)x^3+2x^2-4x-1/(x-2)
x的取值范围是[1,3]且x≠2