Invert Trigonometric System of Equations In my research, I have a system of trig equations that I would like to invert. I know $\alpha$ and $\beta$ ($\theta$ is a constant) in the following system.
$$\cos \alpha  = \cos \theta \cos x + \sin \theta \sin x \cos y$$
$$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$
I would like to find an inversion in the form:
$$\text{trig}(x) =\; ...$$
$$\text{trig}(y) = \;...$$
How can this be done?

$$\cos(\alpha)  = \cos (\theta) \cos( x) + \sin (\theta) \sin (x) \cos( y)\tag 1$$
$$\tan (\beta) = \frac{\cos (x) +\cos (\theta) \cos (\alpha)}{\sin (\theta) \sin (x) \sin (y)}\tag 2$$
From $(1)$
$$\cos(y)=\csc (\theta ) \csc (x) (\cos (\alpha )-\cos (\theta ) \cos (x))\tag 3$$
From $(2)$
$$\sin(y)=\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta )
   \cot (x)\tag 4$$
Using $\cos^2(y)+\sin^2(y)=1$, we end with
$$1=(\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta )
   \cot (x))^2+(\cot (\theta ) \cot (x)-\cos (\alpha ) \csc (\theta ) \csc (x))^2$$
Now, using the tangent half-angle substitution $x=2 \tan ^{-1}(t)$, we end with a quadratic in $t^2$
$$a + b t^2+c t^4 =0$$ where
$$a=16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha )
   \cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3
   \cot ^2(\theta )-3$$
$$b=-2 \cos (2 (\alpha -\beta ))-2 \cos (2 (\alpha +\beta ))-16 \sin ^2(\alpha ) \csc
   ^2(\theta )-$$ $$4 \cos (2 \alpha )+4 \cos (2 \beta )-12$$
$$c=-16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha )
   \cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3
   \cot ^2(\theta )-3$$
Applied to the case use by @Jean Marie $\alpha=\theta=\frac\pi 4 $ and $\beta=\tan ^{-1}(2))$, this leads to
$$132 t^4-152 t^2+36=0$$ giving the four roots
$$t=\pm  \frac{1}{\sqrt{3}} \qquad \text{and} \qquad t=\pm \frac{3}{\sqrt{11}} $$ that is to say
$$x=\pm \frac \pi 3\approx \pm1.04720 \qquad \text{and} \qquad x=\pm2 \tan ^{-1}\left(\frac{3}{\sqrt{11}}\right)\approx 1.47063$$ and then the corresponding $y$'s using $(3)$ or $(4)$.
