calculus - continued square root function

Solutions of following eqution
$$x=\sqrt{2+{\sqrt{2-{\sqrt{2+{\sqrt{2-x}}}}}}}$$
is $\frac{1+\sqrt5}{2}$.
This is solution of
$$x=\sqrt{2+{\sqrt{2-x}}}$$

Does all of this type equation(repeating same shape) always have same solutions like this?
Can you explain why?

Answer

Note the following identities: $$2+2\sin(\theta) = 4\sin^2(\theta/2+\pi/4)$$ $$2-2\sin(\theta) = 4\sin^2(\pi/4-\theta/2)$$

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

where $x = 2\sin(\theta)$.

Using the above identities, it must be that

$$2\sin(\theta)=\sqrt{2+{\sqrt{2-{\sqrt{2+2\sin(\pi/4-\theta/2)}}}}}$$

$$2\sin(\theta)=\sqrt{2+{\sqrt{2-2\sin(3\pi/8-\theta/4)}}}$$

$$2\sin(\theta)=\sqrt{2+2\sin(\theta/8+\pi/16)}$$

$$\sin(\theta) = \sin(\theta/16 + 9\pi/32)$$

$$\theta = 3\pi/10$$

Note that $$2\sin\left(\dfrac{3\pi}{10}\right) = \dfrac{1+\sqrt{5}}{2}.$$

I would conjecture that, yes, equations in that general form will have a solution in the form of the sine or cosine of some rational multiple of $\pi$.

It is worth noting that $\sin\!\left(\!\dfrac{p}{q}\pi\!\right)$ will be algebraic per this thread.