先化为f（x）=-sin（x+π/3）+ acos(x+π/3)
然后化为同名方程：f（x）=-【（√（1+a²））×sin（x+π/3-β）】其中tanβ=-a
因为其对称轴方程为x=π/2所以sin（π/2+π/3-β）=±1
∴π/2+π/3-β=π/2+kπ
β=π/3-kπ
∴a=-tanβ=-tan（π/3-kπ）=-tan（π/3）=-√3
函数f(x)= cos(x-π/6) + acos(x+π/3) (a
1个回答
相关问题
-
设函数f(x)=sin3x+acos2x 满足f(π/6 -x)=f(π/6 +x )则a等于
-
f(x)=Acos((x/4+π/6) f(π/3)=√2
-
f(x)=cos(x+π/6）cos(x-2π/3)+sin(x+π/6)sin(x+π/3) 求函数F（X）的单调递减
-
已知函数f(x)=cos(x+π/6)/sin(π/3+x)
-
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/
-
若函数f{x}=3cos{ωx+Ψ}对于任意X都有f{π/6-x}=f{π/6+x},则f{π
-
已知函数f(x)=sinx^2-acos(π/2-x)+1/2a,x∈[-π/6,5π/6],记f（x）的最小值为g（a
-
已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
-
求函数f(x)=sin(2x+π/3)+2cos^2(2x+π/3),x∈(-π/3,π/6]的值域
-
求函数f(x)=3sin(x+π/3)+cos(x+π/6)周期、最值