We have, 1sinπn−1sin3πn=1sin2πn⇒sin3πn−sinπnsinπnsin3πn=1sin2πn⇒2cos2πnsinπnsinπnsin3πn=1sin2πn⇒2sin2πncos2πn=sin3πn⇒sin4πn−sin3πn=0⇒2cos7πnsinπn=0⇒cos7πn=0 or sinπ2n=0⇒7π2n=(2k+1)sinπ2) or π2n=2kπ where kϵZ⇒n=72k+1 or n=14k
(n=14k not possible for any integral value of k)
As n>3, for k=0, we get n=7