设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/

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设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/3)的值
由（1）你能得到什么结论?

题目是 f(x=sin(x+π/3)+2sin(x+π/3)-√3cos(2π/3-x) !
再问：设函数f(x)=sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x)
再答： f(x)=sin(x+π/3)+2sin(x+π/3)-√3cos(π-π/3-x) =3sin(x+π/3)-√3cos[π-(x+π/3)]. =3sin(x+π/3)+√3cos(x+π/3). =2√3[(√3/2si(x+π/3)+(1/2)cos(x+π/3)] . =2√3[sin(π/3)*sin(x+π/3)+cos(π/3)*cos(x+π/3)] =2√3{cos(x+π/3-π/3). =2√3cosx. f(π/6)=2√3cos(π/6) =2√3*(√3/2). =3. f(π/3)=2√3cos(π/3). =2√3*(1/2). =√3.
再问：由（1）你能得到什么结论？
再答：我没有发现有某种通用规律的结论，请把你的结论说出来，知识共享，好吗？ f(π/6)=[f(π/3)]^2 ，或f(π/3)=f(π/6)/√3. ---->这只是本题的个例，不具普遍性。
再问： f(x)=0

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