Below is the signal :
$$ y[n] = \sin\left( \frac{6\pi}{7} n + 1 \right) $$
According to me the Fundamental period is $7/3$ but is the signal periodic? I think it should satisfy this $\sin(6(\pi/7)n + 1 ) = \sin(6(\pi/7)n + 1 + 7/3 )$ , do I have to draw the signal to see if it is periodic or is there a formula I can multiply/add/divide some values of the equation and find out? To make it short, how is a question like this solved normally...
A $\sin$ signal has a period of $2\pi$.
Therefore you wanna know what $n$ has to be for $\frac{6\pi}7n$ to equal $2\pi.$
$\frac{6\pi}7n=2\pi\Leftrightarrow n=2\pi\frac{7}{6\pi}=\frac{14}6=\frac{7}3$
Therefore the period is $\frac{7}3$
You can check that way :
$y[n+\frac{7}3]=\sin\left(\frac{6\pi}7\cdot(n+\frac{7}3)+1\right)=\sin\left(\frac{6\pi}7n+2\pi+1\right)=\sin\left(\frac{6\pi}7n+1\right)=y[n]$
It works !