函数f(x)是周期兀的偶函数,且当x=【0,兀】时,f(x)=√3tanx-1,则f(8兀/3)的值为

∵函数f(x)是周期兀的偶函数,∴f(x+π)=f(x)∴f(8π/3)=f(-π/3),又因为f(x)是偶函数,所以f(-π/3)=f(π/3)π/3属于[0,π],代入f(x)=√3tanx-1得：f(8π/3)=f(π/3)=3-1=2说明：其实这道题目有问题,因为还有一个答案...