I mean $$\cos \left(\frac{a+2 x}{\frac{5 a^2}{2}+2 a x+2 x^2}\right)=\cos \left(\frac{2 x-a}{\frac{5 a^2}{2}-2 a x+2 x^2}\right) \wedge a\geq \frac{1}{2 \pi } $$ over the reals which depends on a real-valued parameter $a$.
I came across it in the Internet, knowing the answer $ \left\{0, \frac {a \sqrt 5} 2, -\frac {a \sqrt 5} 2\right\}$. My Mathematica is running for hours without any output, solving it.
Using $$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$ we get $$\sin\left(\frac{a+2 x}{5 a^2+4 a x+4 x^2}+\frac{2 x-a}{5 a^2-4 a x+4 x^2}\right)=0\tag1$$or$$\sin\left(\frac{a+2 x}{5 a^2+4 a x+4 x^2}-\frac{2 x-a}{5 a^2-4 a x+4 x^2}\right)=0\tag2$$
First, let us solve $(1)$.
We have $$(1)\implies \frac{a+2 x}{5 a^2+4 a x+4 x^2}+\frac{2 x-a}{5 a^2-4 a x+4 x^2}=k\pi$$ where $k\in\mathbb Z$.
Here, let $$p(x):=\frac{a+2 x}{5 a^2+4 a x+4 x^2}$$ Then, we get $$p'(x)=\frac{-2(2x-a)(2x+3a)}{(4x^2+4ax+5a^2)^2}$$ from which we have $$-\frac{1}{4a}=p\left(-\frac 32a\right)\le p(x)\le p\left(\frac a2\right)=\frac{1}{4a}\tag3$$
Also, let $$q(x):=\frac{2 x-a}{5 a^2-4 a x+4 x^2}$$ Then, we get$$q'(x)=\frac{-2(2x+a)(2x-3a)}{(5a^2-4ax+4x^2)^2}$$ from which we have $$-\frac{1}{4a}=q\left(-\frac a2\right)\le q(x)\le q\left(\frac 32a\right)=\frac{1}{4a}\tag4$$
From $(3)(4)$, we get $$-\pi \le -\frac{1}{2a}\le p(x)+q(x)\le\frac{1}{2a}\le \pi\tag{$\star$}$$ implying $$-\pi\le k\pi\le \pi$$ from which $$ -1\le k\le 1$$ follows.
If $k=0$, then we get $$\begin{align}&\frac{a+2 x}{5 a^2+4 a x+4 x^2}+\frac{2 x-a}{5 a^2-4 a x+4 x^2}=0 \\\\&\implies (a+2x)(5a^2-4ax+4x^2)+(2x-a)(5a^2+4ax+4x^2)=0 \\\\&\implies a(-8ax)+2x(10a^2+8x^2)=0 \\\\&\implies x(4x^2+3a^2)=0 \\\\&\implies x=0\end{align}$$
If $k=1$, then we have $$p(x)+q(x)=\pi$$ This is possible only when $$p(x)+q(x)=\frac{1}{2a}=\pi$$ by $(\star)$.
From $(3)(4)$, we already have $$p(x)\le\frac{1}{4a}\qquad\text{and}\qquad q(x)\le\frac{1}{4a}$$ Considering these, we see that $p(x)+q(x)=\frac{1}{2a}$ is possible only when $$p(x)=q(x)=\frac{1}{4a}\tag5$$ $p(x)=\frac{1}{4a}$ is attained only when $x=\frac a2$ while $q(x)=\frac{1}{4a}$ is attained only when $x=\frac 32a$. So, we see that $(5)$ is impossible.
If $k=-1$, we have to have $$p(x)=q(x)=-\frac{1}{4a}\tag6$$$p(x)=-\frac{1}{4a}$ is attained only when $x=-\frac 32a$ while $q(x)=-\frac{1}{4a}$ is attained only when $x=-\frac a2$. So, we see that $(6)$ is impossible.
Therefore, solving $(1)$ gives $x=0$ which is sufficient.
Next, let us solve $(2)$.
For $k\in\mathbb Z$, $$(2)\implies p(x)-q(x)=k\pi$$ From $(3)(4)$, we have $$-\pi \le -\frac{1}{2a}\le p(x)-q(x)\le\frac{1}{2a}\le \pi$$ from which $$-1\le k\le 1$$follows.
If $k=0$, then we have $$\begin{align}&\frac{a+2 x}{5 a^2+4 a x+4 x^2}-\frac{2 x-a}{5 a^2-4 a x+4 x^2}=0 \\\\&\implies (a+2x)(5a^2-4ax+4x^2)-(2x-a)(5a^2+4ax+4x^2)=0 \\\\&\implies a(10a^2+8x^2)+2x(-8ax)=0 \\\\&\implies 4x^2-5a^2=0 \\\\&\implies x=\pm\frac{\sqrt 5}{2}a\end{align}$$
If $k=1$, we have to have $$p(x)=\frac{1}{4a}\qquad\text{and}\qquad q(x)=-\frac{1}{4a}\tag7$$ $p(x)=\frac{1}{4a}$ is attained only when $x=\frac a2$ while $q(x)=-\frac{1}{4a}$ is attained only when $x=-\frac a2$. So, we see that $(7)$ is impossible.
If $k=-1$, we have to have $$p(x)=-\frac{1}{4a}\qquad\text{and}\qquad q(x)=\frac{1}{4a}\tag8$$$p(x)=-\frac{1}{4a}$ is attained only when $x=-\frac 32a$ while $q(x)=\frac{1}{4a}$ is attained only when $x=\frac 32a$. So, we see that $(8)$ is impossible.
Therefore, solving $(2)$ gives $x=\pm\frac{\sqrt 5}{2}a$ which are sufficient.
Conclusion :
$$\text{$x=0,\ \pm\frac{\sqrt 5}{2}a\ $ are the only solutions.}$$