Cezar wrote: Find all pairs $(a,b)$ of real numbers, so that $\sin(2009x)+\sin(ax)+\sin(bx)=0$ holds for any $x\in \mathbf {R}$. If $a=0$, we get $b=-2009$ If $b=0$, we get $a=-2009$ If $ab\ne 0$ : Take double derivative of equation and again double derivative. We get then a $3\times 3$ linear system whose determinant must be $0$, which means $a=\pm 2009$ or $b=\pm 2009$ And trivially, none of these possibilities is a solution (remember $ab\ne 0$) Hence the result : $\boxed{(a,b)\in\{(0,-2009),(-2009,0)\}}$