[[complex]] Solutions of $x=\sqrt{\sin{x}}$ I wanted to make one of those cool infinite recursive definitions for myself, and I chose one that I thought looked cool: $x=\sqrt{\sin{x}}=\sqrt{\sin{\sqrt{\sin{\sqrt{\sin{...}}}}}}$ for no other reason than because I thought it looked cool. Using my method* of finding solutions for these; start with a number e.g. 1, take sqrt(sin(1)), take sqrt(sin(that)) alot of times using Answer button until you find a good guess, then algebraically confirm said guess; I got as far as $0.8767262154$ on my TI-83 Plus and $0.87672621539$ on the calculator you get on http://www.google.com/search?q=calc and also don't forget $0$ courtesy of Mac Grapher.app,

wolfram alpha roasted all with a massive $0.876726215395062445972118643142$

What my real question is, is how would I calculate (not just compute) or solve for all the real/complex solutions of this? Also, my complex question no pun intended is, wolframalpha also gave me $x=\frac{\sqrt{i\left(e^{-ix}-e^{ix}\right)}}{\sqrt{2}}$, how did they get this?[[What are the complex solutions of this equation?]]

*I discovered this for myself when testing that $φ=1+\frac{1}{φ}$ but am aware that it may have already occured to people who know about Mandelbrot set and higher degree 2-dimensional polynomial equations for which algebraic methods of calculation hav yet to be developed
For the complex looking thing,

Remember that
$$e^{ix}=\cos x+\sin x$$
$$e^{-ix}=\cos x - \sin x$$

So $$e^{-ix} -e^{ix}=-2\sin x $$

Hence

\begin{align}
x&amp;=\sqrt{\sin x}\\
&amp;=\sqrt{\frac{e^{-ix} -e^{-ix}}{-2}}\\
&amp;=\frac{\sqrt{i(e^{-ix} -e^{ix})}}{\sqrt2}
\end{align}
So cmarangu as for your complex solutions there are ~3.

Theres 0i.

And then theres one around 0.5231i

And then theres one around 4.3069i

And don't forget ∞i.

I found these solutions using Mac Grapher app I said y=x and found intersections with y=sqrt(sin(ix)) if ya know what I mean.

In order to find the exact values for these you may have to do some weird stuff with e^(ix) and ya.