Do you know basic trig 1

we know that the only int values of sin(n) over real n are $\sin(n)=-1,0,1$ for each we have $n=-\dfrac{\pi}{2}+2k\pi,\boxed{0}+2k\pi,\dfrac{\pi}{2}+2k\pi$ if you got tricked by thinking the question asked degrees, remember if the degree symbol is not given it means the argument is in radians.

The only integral values sin(n) can take are 0,-1 and 1(since range of sin function is [-1,1]),n=0 is the only integral solution for which sin(n) is also a integer.(sin0=0∈Z)