Solve the following equation: $\frac{1}{x^2}+\frac{1}{(4-\sqrt{3}x)^2}=1$ 
Solve the following equation: $$\frac{1}{x^2}+\frac{1}{(4-\sqrt{3}x)^2}=1$$

I know it's from a Math Olympiad but I don't know which and I couldn't find it on the internet. Expanding everything doesn't work (it leads you to a 4th degree equation with no rational roots). I don't know what else to try, I can't see any useful substitution.
Any help would be appreciated, thanks.
 A: If you do a substitution $t = x \sqrt 3$, you get
$$
\frac 3 {t^2} + \frac 1{(4 - t)^2} = 1 \implies t^4 - 8t^3 + 12 t^2 + 24t - 48=0
$$
You can check that $t = 2$ is a solution, so $P_4(t) = (t-2)P_3(t)$, therefore $x = \frac 2{\sqrt 3}$ is a solution of the initial equations.
A: It seems like you're right, Kasper. A trigonometric substituition would solve the equation completely. I'll write the solution I arrived at:
Finding inequalities is helpful to motivate the right subtituition. We know that both $\frac{1}{x^2}$ and $\frac{1}{(4-\sqrt{3}x)^2}$ are positive, so we can say that:
$$\frac{1}{x^2}<1\Leftrightarrow \left|\frac{1}{x}\right|<1\Leftrightarrow -1<\frac{1}{x}<1$$
And analogously $\displaystyle-1<\frac{1}{4-\sqrt{3}x}<1$. The substituition is starting to become obvious. One last hint is that we're trying to find two squares whose sum equals $1$. This immediately reminds us of the identity $\sin^2 a+\cos^2 a=1$. That's enough to motivate the substituition (WLOG):
$$\frac{1}{x}=\sin a \,\,\text{ and }\,\,\frac{1}{4-\sqrt{3}x}=\cos a$$
Which will give us:
$$\begin{align}
&\frac{1}{\displaystyle4-\frac{\sqrt{3}}{\sin a}}=\cos a \Leftrightarrow \\\\
&\frac{1}{\cos a}+\frac{\sqrt{3}}{\sin a} = 4 \Leftrightarrow \\\\
&\frac{1}{2}\sin a + \frac{\sqrt{3}}{2}\cos a = 2\sin a\cos a \Leftrightarrow \\\\
&\sin(60°+a)=\sin(2a)
\end{align}$$
Solving and substituting back, the solutions are: $$x_1=\csc60°=\frac{2}{\sqrt{3}},\;x_2=\csc 40°,\;x_3=\csc20°,\;x_4=-\sec10°.\;\;\blacksquare$$
A: Let $u = \sqrt{3}x$, and $v = 4 - \sqrt{3}x$, then:
$u + v = 4$, and $\dfrac{3}{u^2} + \dfrac{1}{v^2} = 1 \to 3v^2 + u^2 = u^2v^2 \to 3v^2 = u^2(v^2 - 1) \to 3v^2 = (4-v)^2(v^2 - 1)$. Observe that $v=2$ is a root to the above equation. From this we can use synthetic division to factor the polynomial and finish it.
A: One solution is "clear" at $\frac{2}{\sqrt{3}}$. I was motivated to look for something like this by trying to write $1$ as the sum of two simple fractions, and the presence of $3$ and $4$. 
It's also halfway between the two vertical asymptotes of $\frac1{x^2}+\frac{1}{\left(4-\sqrt{3}x\right)^2}$, and in a sketch of that function (which clearly is never negative) it was natural to want to see what the output was at that midpoint.
