How can I solve this system, to find $\alpha$, $\beta$ and $\gamma$?
$$\begin{aligned} x & = \cos\alpha\sin\beta\sin\gamma - \sin\alpha\cos\gamma \\ y & = \sin\alpha\sin\beta\sin\gamma + \cos\alpha\cos\gamma \\ z & = \cos\beta\sin\gamma \end{aligned} $$
In other words, I want to rewrite it like:
$$\begin{aligned} \alpha & = \dots \\ \beta & = \dots \\ \gamma & = \dots \end{aligned} $$
($x$, $y$ and $z$ are known)
Is there any online tool that could do it for me? Wolfram alpha doesn't handle.
I think that I had a way to solve the problem.
For brevity, i wrote the equations as $$c(\alpha ) s(\beta ) s(\gamma )-c(\gamma ) s(\alpha )-x=0 \tag 1$$ $$c(\alpha ) c(\gamma )+s(\alpha ) s(\beta ) s(\gamma )-y=0 \tag 2$$ $$c(\beta ) s(\gamma )-z=0 \tag 3$$ $$c(\alpha )^2+s(\alpha )^2-1=0 \tag 4$$ $$c(\beta )^2+s(\beta )^2-1=0 \tag 5$$ $$c(\gamma )^2+s(\gamma )^2-1=0 \tag 6$$ where $c(t)=\cos(t)$ and $s(t)=\sin(t)$.
May be obvious : squaring $(1)$, $(2)$, $(3)$ and summing gives $x^2+y^2+z^2=1$; if this is not the case, end of the problem.
Now, I started eliminating one at the time the variables. This gives $$s(\beta )=\frac{c(\gamma ) s(\alpha )+x}{c(\alpha ) s(\gamma )}$$ $$c(\gamma )=y c(\alpha )-x s(\alpha )$$ $$c(\beta )=\frac{z}{s(\gamma )}$$ $$s(\gamma )=\pm\sqrt{2 x y c(\alpha ) s(\alpha )+x^2 c(\alpha )^2+y^2 s(\alpha )^2+z^2}$$and the final equation is $$x(1+c(2\alpha))+y s(2\alpha))=0$$ that is say $$x (1+\cos (2 \alpha ))+y \sin (2 \alpha )=0$$ Excluding the trial $\alpha=\pm \frac \pi 2$, this leads to $$\alpha=\pm \cos ^{-1}\left(\pm\frac{y}{\sqrt{x^2+y^2}}\right)$$ Going backward, we have the solutions for $\beta$ and $\gamma$.