How to use double angle identities to find length 'd'?

Question

Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.

A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:

Edit: Note that triangle ACB is not a right angled triangle. Should now read: Note that triangle ACD is not a right angled triangle.

By using $$\cos 2x=2\cos^2 x-1$$ My working is: $$\cos x=\frac d3$$ $$\cos 2x=\frac d3$$ therefore: $$\frac d3=2(\frac d3)^2-1$$

Now becomes solvable as a quadratic equation

Answer 1

To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$

Answer 2

$\cos(x)\neq \frac{2}{3}$ because the angle $\widehat{ACD} \neq \frac{\pi}{2}$.

Answer 3

I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"

Using the law of cosines in triangles $ABC$ and $ACD$ we get $$c^2=2^2+3^2-2\times 2\times 3 \cos x$$ $$b^2=d^2+2^2-4d\cos x$$ so that $$\cos x=\dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-\dfrac{4d(13-c^2)}{12}$$ The last equation is quadratic in $d$ and easy to solve.