问题标题：
已知f(x)=ax^3+bx+sin2x+cosx+1,且f(π/3)=3,求f(-π/3)
问题描述：
已知f(x)=ax^3+bx+sin2x+cosx+1,且f(π/3)=3,求f(-π/3)
唐紫英回答：
f(π/3)=a(π/3)^3+b(π/3)+sin(2π/3)+cos(π/3)+1=3a(π/3)^3+b(π/3)=(3-√3)/2f(-π/3)=-a(π/3)^3-b(π/3)-sin(-2π/3)+cos(-π/3)+1=-[a(π/3)^3+b(π/3)]+√3/2+1/2+1=-(3-√3)/2+√3/2+1/2+1=√3
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