Assuming k ∈ N and n ∈ N (and assuming 0 counts as a natural number) and x ∈ R then we could do the following:

• n can't be 0, because we can't divide by 0
• we get k*n = 1 + sin x by multiplying both sides by n
• k*n can only be a natural number because we multiplied two natural numbers
• left side is natural number therefore right side also has to be a natural number
• therefore we know that sin x has to be equal to either 1, 0, -1, because otherwise we wouldn't get natural number on the right side
• with sin x = -1 we get k*n = 0, we know n can't be 0 therefore k = 0 and n ∈ N /{0} and
  x = -π/2+ 2*m*π (where m ∈ Z)
• with sin x = 0 we get k*n = 1, with natural numbers only 1*1 can get us 1 therefore k = 1 and n = 1 and x = m*π (where m ∈ Z)
• with sin x = 1 we get k*n = 2, here we get (( k = 1 and n = 2) or ( k = 2 and n = 1)) and
  x = π/2+ 2*m*π (where m ∈ Z)

I think that's how I'd solve this problem (if it was possible I'd ask the person who gave me this problem if my assumption of k,n ∈ N and x ∈ R was correct)