For there to be a vertical asymptote, the denominator bust be 0.
Here, we have #(x-4)sin(\pi+(3\pi)/x)# as the denominator. As long as one of these equals 0, the entire denominator will equal 0.
An asymptote will appear at #x=4# because if we put out #x# value in, we get #0sin(\pi+(3\pi)/4)=0#.
A second asymptote will appear at #x=0# because in the #sin# we have a fraction with #x# on the bottom, if this equals 0 then there can be no #y# value.
The final asymptotes will appear when #x=3/(n-1), ninRR#, this is because for #sin\theta# to equal 0, #\theta# must equal #n\pi, ninRR#.
So, we have #n\pi=\pi+(3\pi)/x#
If we rearrange to make #x# the subject, we get #x=(3\pi)/(n\pi-\pi)=3/(n-1)#
By putting this into #sin(\pi+(3\pi)/x)#, we get #sin(\pi+(3\pi)/(3/(n-1)))-=sin(\pi+(3\pi(n-1))/3-=sin(\pi+\pi(n-1))-=sin(\pi+n\pi-\pi)-=sin(n\pi)#
