If the function \(f(x)=a\sin x+\frac{1}{3}\sin 3x\) has maximum value at \(x=\frac{\pi}{3}\) then the value of \(a\) is

\(3\) \(\frac{1}{3}\) \(2\) \(\frac{1}{2}\)

\(2\)

\(f^{\prime}(x)=a\cos x+\cos 3x\hspace{9 cm}\) Given, \(f^{\prime}\left(\frac{\pi}{3}\right)=0\) So, \(a=2\)