已知函数f(X)=2sin(x-π/3)cos(x-π/3)+2根号3cos^2(x-π/3)-根号3,若函数y=f(2x)-a在区间
题目
已知函数f(X)=2sin(x-π/3)cos(x-π/3)+2根号3cos^2(x-π/3)-根号3,若函数y=f(2x)-a在区间
[0,π/4]上恰有两个零点x1,x2,求tan(x1+x2)的值
答案
f(x)=2sin(x-π/3)cos(x-π/3)+2√3cos^2(x-π/3)-√3 =sin(2x-2π/3)+√3cos(2x-2π/3) =2[sin(2x-2π/3)cosπ/3+cos(2x-2π/3sinπ/3] =2sin(2x-2π/3+π/3)+1-√3 =2sin(2x-π/3) 故 f(2x)-a=2sin...
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