<s>[INST] solving $\sin(5x)+\sin(x)=0$ using the sum formulas so symbolab says its just
$$2 \sin(3x) \cos(2x)$$
But after applying the angle sum formula I get
$$\sin(5x)+\sin(x)=\sin(3x+2x)+\sin(x)=\sin(3x)\cos(2x)+\sin(2x)\cos(3x)+\sin(x)$$
How does this reduce more?
 [/INST] Using the compound angle formula, $$
\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2},
$$
we have
$$
\begin{array}{l}
\sin 5 x+\sin x=0 \\
2 \sin 3 x \cos 2 x=0 \\
\sin 3 x=0 \text { or } \cos 2 x=0 \\
\displaystyle 3 x=n \pi \text { or } 2 x=n \pi+\frac{\pi}{2} \\
\displaystyle x=\frac{n \pi}{3} \text { or } \frac{(2 n+1) \pi}{4},
\end{array}
$$
where $n\in Z.$
</s>