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(A) For $n = 2$,the function is $f(x) = \frac{\sin(2x)}{\sin(x/2)}$.Using the identity $\sin(2	heta) = 2\sin(	heta)\cos(	heta)$,we have $\sin(2x) =

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$2$

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(A) For $n = 2$,the function is $f(x) = \frac{\sin(2x)}{\sin(x/2)}$.Using the identity $\sin(2	heta) = 2\sin(	heta)\cos(	heta)$,we have $\sin(2x) = 2\sin(x)\cos(x) = 2(2\sin(x/2)\cos(x/2))\cos(x) = 4\sin(x/2)\cos(x/2)\cos(x)$.Thus,$f(x) = \frac{4\sin(x/2)\cos(x/2)\cos(x)}{\sin(x/2)} = 4\cos(x/2)\cos(x)$.The period of $\cos(x)$ is $2\pi$ and the period of $\cos(x/2)$ is $\frac{2\pi}{1/2} = 4\pi$.The period of the product of two functions is the least common multiple of their individual periods. The $LCM$ of $2\pi$ and $4\pi$ is $4\pi$.For $n = 3$,$f(x) = \frac{\sin(3x)}{\sin(x/3)}$. The period of $\sin(3x)$ is $2\pi/3$ and the period of $\sin(x/3)$ is $6\pi$. The period of $f(x)$ is $LCM(2\pi/3, 6\pi) = 6\pi 
eq 4\pi$.Similarly,for other values of $n$,the period will not be $4\pi$.Therefore,the correct value is $n = 2$.

The value of $n \in \mathbb{Z}$ for which the function $f(x) = \frac{\sin(nx)}{\sin(x/n)}$ has $4\pi$ as its period,is

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