f(x)=Asinx+cosx
=√(1+A^2)*[sinx*A/√(1+A^2)+cosx*1/√(1+A^2)]
=√(1+A^2)*sin(x+Q),(此時(shí)cosQ=A/√(1+A^2),sinQ=1/√(1+A^2),)
而,Y=√(1+A^2)*sin(x+Q),
此函數(shù)周期為T(mén)=2π,
對(duì)稱軸方程為
X+Q=π/2,而,X=π/3,則有
π/3+Q=π/2.
Q=π/6.
而,cosQ=A/√(1+A^2)=√3/2.兩邊平方得,
4A^2=3A^2+3,
A^2=3,
A1=√3,A2=-√3.
即,A的值為:A1=√3,A2=-√3.