Is signal periodic? What is the period? 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: 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 !
A: Signal is periodic with period 7, not 7/3! Because we are talking about discrete case not continuous case. Here 'N' which is time period of discrete signal is always an integer.
Because 'n' should be integer angular frequency(w) = 6π/7
So time period N = 2πm/(6π/7) = 7m/3
But N is always an integer in discrete case, hence we would multiply it with m = 3. 
Therefore, N = 7
'm' denoted the number of cycles continuous signal   has to repeat to set 1 period of discrete case.
A: A sinusoidal signal (such as $\sin$) is indeed always periodic.
The general form of a sine wave is:
$f(t) = A\sin(2\pi f t + \phi)$
Where $\phi$ (greek letter phi) is the phase shift, $f$ is the frequency (in Hertz, or Hz $=\frac{1}{s} = s^{-1}$) and $A$ is the amplitude of the wave.  
You can use the formula $\omega = 2\pi f$ where $\omega$ is the angular frequency (in $\frac{\text{rad}}{\text{s}}$), with the formula $T = \frac{1}{f}$ where $T$ is the period of the signal (in seconds, $s$).
The (general form) formula then becomes:
$f(t) = A\sin(\omega t + \phi)$ 
In your case (to make your formula fit the general form) you can say that $t=n$, $A=1$, $\phi = 1$ and the angular frequency $\omega = \frac{6\pi}{7}$, which means that the frequency is:
$f = \frac{\omega}{2\pi}=\frac{6\pi}{7}\frac{1}{2\pi}=\frac{3}{7}$.
Now you can find out the period as follows:
$T = \frac{1}{f} = \frac{1}{\frac{3}{7}}=\frac{7}{3}$
It's worthy to note that the term $\phi$ in the equation only offsets the wave (i.e. if $\phi \neq 2\pi k, k=[0,1,2,...]$ then the wave is offset by $\phi$) but does not affect its period/frequency.
