f(x)=Asinx+cosx
=√(1+A^2)*[sinx*A/√(1+A^2)+cosx*1/√(1+A^2)]
=√(1+A^2)*sin(x+Q),(此时cosQ=A/√(1+A^2),sinQ=1/√(1+A^2),)
而,Y=√(1+A^2)*sin(x+Q),
此函数周期为T=2π,
对称轴方程为
X+Q=π/2,而,X=π/3,则有
π/3+Q=π/2.
Q=π/6.
而,cosQ=A/√(1+A^2)=√3/2.两边平方得,
4A^2=3A^2+3,
A^2=3,
A1=√3,A2=-√3.
即,A的值为:A1=√3,A2=-√3.