Two nested radicals with similar result Sorry for my last post it's my bad .So I ask to this (the true ^^)nested radical :
$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{\cdots }}}}}}}}=\frac{1+\sqrt{5}+\sqrt{30-6\sqrt{5}}}{4}$$ 
The period is $4$ and the related equation is :
$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$$
There is a big similarity with this How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$
Following the answer of Tito Piezas III can solve this nested radical (with $2$).   
My question: 
Can someone explain these similarities between these two nested radicals ?
Any helps is highly appreciated .
Thanks a lot .

We can solve the equation $\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$ as follows.
Since $2+x\ge 0$, we have $x\ge -2$. Since $\sqrt{2+x}\le 2$, $x\le 2$. So we can put $x=2\cos 32\alpha$ for some $0\le\alpha\le \tfrac{\pi}{32}$. Then   
$$\sqrt{2+x}=\sqrt{2+2\cos 32\alpha}=2\cos 16\alpha$$
$$\sqrt{2-2\cos 16\alpha}=2\sin 8\alpha$$
$$\sqrt{2-2\sin 8\alpha}=\sqrt{2-2\cos\left(\frac{\pi}2- 8\alpha\right)}=2\sin\left(\frac{\pi}4- 4\alpha\right)$$
$$\sqrt{2+2\sin\left(\frac{\pi}4-4\alpha\right)}=\sqrt{2+2\cos\left(\frac{\pi}4+4\alpha\right)}=2\cos\left(\frac{\pi}8+2\alpha\right)$$ 
So $$2\cos\left(\frac{\pi}8+2\alpha\right)=2\cos 32\alpha $$
$$\sin\left(17\alpha+\frac{\pi}{16}\right)\sin\left(15\alpha-\frac{\pi}{16}\right)=0$$ 
$$17\alpha+\frac{\pi}{16}=\pi n\mbox{ or }15\alpha-\frac{\pi}{16}=\pi n,\,\, n\in\Bbb Z$$ 
For the range $0\le\alpha\le \tfrac{\pi}{32}$ fits only $15\alpha-\frac{\pi}{16}=0$, so $\alpha=\frac{\pi}{15\cdot 16}$ and 
$$x=2\cos 32\alpha=2\cos\frac{2\pi}{15}.$$
