The problem has multiple solutions. Most of the participants who solved the problem chose f(x) = x and g(x) = -x and made use of the fact that sine is an odd function. For so chosen f and g,

Another solution is given by two piecewise-defined functions;

	f(x) = 0 if x ≤ 0 1 otherwise
	g(x) =  0 if x ≤ 1 -1 otherwise

Indeed,

sin(g(x)) =    0 if x ≤ 1 -sin(1) otherwise

so that f(sin(g(x))) = 0 identically. On the other hand,

sin(f(x)) =  0 if x ≤ 0 sin(1) otherwise

and, since sin(1) < 1, g(sin(1)) = 0, making g(sin(f(x)) = 0, for all x.