A false nested radical of Ramanujan and a true nested radical

Question

On the wikipedia page I can see a nested radical by Ramanujan :

$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$$

Wolfram alpha says it's false

So I propose another one wich is true see here we have :

$$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{\cdots}}}}}}}=2\cos\Big(\frac{\pi}{9}\Big)$$

My question :

Can someone give me some steps to solve :

$$\sqrt{2+\sqrt{2+\sqrt{2-x}}}=x$$

I know furthermore that is related to a cubic .

Any helps is highly appreciated .

Thanks a lot for all your contributions.

Ps: Can someone correct the wikipedia page and add the nested radical with $2$?

Answer 1

Solving cyclic infinite nested square roots of 2 in the form $$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-...}}}}}}$$ Refer here Put $x = \sqrt{2+\sqrt{2+\sqrt{2-x}}}$ where $x = 2\cos\theta$

now the equation becomes easy to simplify

$2\cos\theta = \sqrt{2+\sqrt{2+\sqrt{2-2\cos\theta}}}$

= $\sqrt{2+\sqrt{2+{2\sin\frac{\theta}{2}}}}$ ...(from Half angle cosine formula for $\sin(\frac{\theta}{2}$)

= $\sqrt{2+\sqrt{2+{2\cos(\frac{\pi}{2}-\frac{\theta}{2})}}}$ = $\sqrt{2+{2\cos(\frac{\pi}{4}-\frac{\theta}{4})}}$ (from Half angle cosine formula for $\cos(\frac{\theta}{2}$)

= ${2\cos(\frac{\pi}{8}-\frac{\theta}{8})}$

now $\theta = \frac{\pi}{8}-\frac{\theta}{8}$ further simplification leads to

$\frac{9\theta}{8} = \frac{\pi}{8}$

$\therefore$ $\theta = \frac{\pi}{9}$ which is $20^\circ$

It is true that simplifying the cyclic infinite nested square roots of 2 $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-...}}}}}}$ is $2\cos\frac{\pi}{9}$

As for as Ramanujan's nested radical is concerned