The following holds

\(\sin(x+1)=\sin(x)\cos(1)+\cos(x)\sin(1)\),

\(\sin(x+2)=\sin(x)\cos(2)+\cos(x)\sin(2)\),

\(\sin(x+3)=\sin(x)\cos(3)+\cos(x)\sin(3)\).

Using these formulae we obtain the following \(a_1\sin(x+1)+a_2\sin(x+2)+a_3\sin(x+3)=\\ \Big(a_1\cos(1)+a_2\cos(2)+a_3\cos(3)\Big)\sin(x)+ \Big(a_1\sin(1)+a_2\sin(2)+a_3\sin(3)\Big)\cos(x)\).

Now because \(\sin(x)\) and \(\cos(x)\) are linear independent (by the solution to the previous variant) the expression is equivalent to the zero-function if and only if

\(a_1\cos(1)+a_2\cos(2)+a_3\cos(3)=0\) and

\(a_1\sin(1)+a_2\sin(2)+a_3\sin(3)=0\).

But this is a homogeneous system of equations, which consists of two equations with three variables, and thus there must be a non-trivial solution to the system.

One particular solution is \((a_1,a_2,a_3)^T=\\\left( \cos(2)\sin(3)-\cos(3)\sin(2), \cos(3)\sin(1)-\cos(1)\sin(3), \cos(1)\sin(2)-\cos(2)\sin(1) \right)^T=\\\left(\sin(1),-\sin(2),\sin(1)\right)^T \)