[tex]\sin(4x)+\sin(6x)=0[/tex]Expand the arguments to include [tex]5x[/tex]:[tex]\sin(5x-x)+\sin(5x+x)=0[/tex][tex](\sin(5x)\cos x-\sin x\cos(5x))+(\sin(5x)\cos x+\sin x\cos(5x))=0[/tex][tex]2\sin(5x)\cos x=0[/tex]Then either[tex]\sin(5x)=0[/tex]or[tex]\cos x=0[/tex]In the first case, use the fact that [tex]\sin0=\sin\pi=0[/tex] and that [tex]\sin x[/tex] has period [tex]2\pi[/tex], so that[tex]\sin(5x)=0\implies\begin{cases}5x=0+2n\pi\\5x=\pi+2n\pi\end{cases}\implies\boxed{x=\dfrac{2n\pi}5\text{ or }x=\dfrac{(2n+1)\pi}5}[/tex]where [tex]n[/tex] is any integer.In the second case, we have [tex]\cos\dfrac\pi2=\cos\dfrac{3\pi}2=0[/tex] and [tex]\cos x[/tex] also has period [tex]2\pi[/tex], so that[tex]\cos x=0\implies\begin{cases}x=\dfrac\pi2+2n\pi\\\\x=\dfrac{3\pi}2+2n\pi\end{cases}\implies\boxed{x=\dfrac{(4n+1)\pi}2\text{ or }\dfrac{(4n+3)\pi}2}[/tex]