Solving $\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0$

Question

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 $x=-1$.

Answer 1

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 $x=-1$.

Answer 2

It must be $$\frac{x}{2(x^2+1)}+\frac{1}{(2x^2+x+1)}=k\pi$$ or $$\frac{x}{2(x^2+1)}+\frac{1}{(2x^2+x+1)}=\frac{(2k+1)\pi}{2}$$