If the period of the function $\cos(nx)\sin(5x/n)$ is $3\pi$ then what should be number of integral values of $n$ ?
My approach : I tried like period of $\cos(nx)$ is $2\pi$/n and $\sin(5x/n)$ is $2\pi n/5$ So the period should be L.C.M of $2\pi$/n and $2\pi n/5$.Which is equal to $2\pi n/\gcd(n,5)$. However after this I'm not being able to proceed. Help please!
You're given
$$\cos nx\cdot\sin\frac{5x}n=\cos(n(x+3\pi))\cdot\sin\frac{5(x+3\pi)}n$$
Using some trigonometry:
$$\cos(n(x+3\pi))\cdot\sin\frac{5(x+3\pi)}n=\cos(nx+3n\pi)\cdot\sin\left(\frac{5x}n+\frac{15\pi}n\right)=$$
$$=\overbrace{(-1)^n}^{=\cos3n\pi}\cos nx\left(\sin\frac{5x}n\cos\frac{15\pi}n+\sin\frac{15\pi}n\cos\frac{5x}n\right)$$
Equalling the last expression to the left side of the first one above, we get:
$$\sin\frac{5x}n=(-1)^n\left(\sin\frac{5x}n\cos\frac{15\pi}n+\sin\frac{15\pi}n\cos\frac{5x}n\right)$$
As this must be true for any value of $\;x\;$ , we can choose say $\;x=0\;$ : $$0=(-1)^n\left(0+\sin\frac{15\pi}n\right)\implies n\in\pm\{1,3,5,15\}$$
and we've reduced the possible choices for $\;n\;$ . Take it from here.
