<s>[INST] Solving $\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0$ 
I'm looking for a more elegant way to find the real solution
$$\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0.$$

Here is my way:
First write the equation as
$$\sin\left(\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+x+2)}\right)\cos\left(\frac{x}{2(x^2+1)}-\frac{1}{2(x^2+x+2)}\right)=0.$$
Then the real solution can be find by
$$\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+x+2)}=0$$
which gives $x=-1$.
 [/INST] I do not know if this is more elegant.
Let $f(x)=\sin x$, $g(x)=\frac{x}{x^2+1}$, and $h(x)=\frac{1}{x^2+x+2}$.
Then your equation becomes
$$f\left(g(x)\right)+f\left(h(x)\right)=0$$
Since $f(x)=\sin x$ is an odd function, 
$$-f\left(h(x)\right)=f\left(-h(x)\right).$$
Then, your equation is that
$$f\left(g(x)\right)=f\left(-h(x)\right).$$
It is known that in the range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ the function $f(x)$ is increasing, and since $-\frac{1}{2}\leq g(x)\leq\frac{1}{2}$ and $0<h(x) \leq \frac{4}{7}$, your equation is equivalent to
$$g(x)=-h(x)\leftrightarrow \frac{x}{x^2+1}=-\frac{1}{x^2+x+2}.$$
Therefore, the real solution comes from
$$x^3+2x^2+2x+1=0,$$
which gives $x=-1$.
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