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?
 A: Use the fact that\begin{align}\require{cancel}\sin(5x)+\sin(x)&=\sin(3x+2x)+\sin(3x-2x)\\&=\sin(3x)\cos(2x)+\cancel{\sin(2x)\cos(3x)}+\\&\qquad+\sin(3x)\cos(2x)-\cancel{\sin(2x)\cos(3x)}\\&=2\sin(3x)\cos(2x).\end{align}
A: Consider the formulas
$$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$
$$\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)$$
Adding we get:
$$\sin(a+b)+\sin(a-b) = 2 \sin(a)\cos(b) \tag{1}$$
Let $u=a+b$ and $v=a-b$, so we have $a = \frac{u+v}{2}$ and $b = \frac{u-v}{2}$, and $(1)$ becomes:
$$\bbox[10px,border:1px solid]{
\sin(u)+\sin(v) = 2 \sin(\tfrac{u+v}{2})\cos(\tfrac{u-v}{2})
}$$
Now use this formula with $u=5x$ and $v=x$
A: 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 $n\in Z.$
