Solving systems of equation with the unknowns nested in trigonometric functions Would be nice, if someone could help me with following systems of equation with the unknowns nested in trigonometric functions.
Find the angles $\alpha_1$ and $\alpha_2$ solving the equations
$$
\begin{align}
x_1&= \sin(\beta_{11}+\alpha_1)+ \sin(\beta_{21}+\alpha_2) \\
x_2&= \sin(\beta_{12}+\alpha_1)+ \sin(\beta_{22}+\alpha_2).
\end{align}
$$
All $\beta_{ij}$, $x_1$, and $x_2$ are known. I tried to apply the idea from 
http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-rcostheta-alphaetc.pdf, but struggled, since the unknowns are nested in the arguments of $\sin$.
A substitution didn’t worked for me, since then we end up with 4 unknowns.
Any help would be very appreciated.
 A: I don't know if this helps, but I could propose a semi-linear approach in order to reduce the problem to jointly solving two quadratic equations. I thought this would be better than nothing. Let's begin:
First, using the angle sum identities, your system of equations could be written as:
$$
x_1 = k_1 \cos(\alpha_1) + k_2 \sin(\alpha_1) + k_3 \cos(\alpha_2) + k_4 \sin(\alpha_2)\\
x_2 = k_5 \cos(\alpha_1) + k_6 \sin(\alpha_1) + k_7 \cos(\alpha_2) + k_8 \sin(\alpha_2)
$$
where 
$$k_1=\sin(\beta_{11}),\, k_2=\cos(\beta_{11}),\,  k_3=\sin(\beta_{21}),\,  k_4=\cos(\beta_{21}),\, \\ k_5=\sin(\beta_{12}),\,  k_6=\cos(\beta_{12}),\,  k_7=\sin(\beta_{22}),\,  k_8=\cos(\beta_{22})$$
We can wrap this to a matrix form:
$$
\begin{bmatrix}
k_1 & k_2 & k_3 & k_4 \\
k_5 & k_6 & k_7 & k_8
\end{bmatrix}
\begin{bmatrix}
a\equiv\cos(\alpha_1)\\
b\equiv\sin(\alpha_1)\\
c\equiv\cos(\alpha_2)\\
d\equiv\sin(\alpha_2)
\end{bmatrix}=\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
$$
Albeit underdetermined, we could still solve this system (e.g. using LU/SVD) to get a particular solution $\mathbf{s}$ and the two null space vectors, $\mathbf{n}_1$ and $\mathbf{n}_2$. The sought solution $\mathbf{q}$ can then be written as a linear combination of these basis vectors:
$$
\begin{equation}
\mathbf{q}=\begin{bmatrix}a\\b\\c\\d\end{bmatrix} = \mathbf{s}+\lambda_1\mathbf{n}_1+\lambda_2\mathbf{n}_2
\end{equation}
$$
The solution is underdetermined because we had only two linear constraints at our disposal. We now make use of the non-linear ones. Notice, the first two rows and the last two rows of $\mathbf{q}$ are related:
$$
a = \cos(\arcsin(b)) = \sqrt{1-b^2}\\
c = \cos(\arcsin(d)) = \sqrt{1-d^2}
$$
The right hand side arises due to the identity $\cos(\arcsin(x))=\sqrt{1-x^2}$. We can then plug this into the solution parameterization to get the two non-linear constraints:
$$
\mathbf{s}^1 + \lambda_1 \mathbf{n}_1^1+ \lambda_2 \mathbf{n}_2^1 = \sqrt{1- (\mathbf{s}^2 + \lambda_1 \mathbf{n}_2^2+ \lambda_2 \mathbf{n}_2^2)^2 }\\
\mathbf{s}^3 + \lambda_1 \mathbf{n}_1^3+ \lambda_2 \mathbf{n}_2^3 = \sqrt{1- (\mathbf{s}^4 + \lambda_1 \mathbf{n}_2^4+ \lambda_2 \mathbf{n}_2^4)^2 }
$$
$\mathbf{x}^i$ denotes the $i^{\text{th}}$ component of the vector $\mathbf{x}$. At this stage as $\mathbf{n}_1,\mathbf{n}_2 \text{ and }\mathbf{s}$ are known, the only unknowns to determine are $\lambda_1$ and $\lambda_2$. Squaring both sides of the equation above gives us two second degree polynomials, which we could solve to get these variables. Unfortunately, although I am not $100\%$ sure, the solution might involve solving a system of polynomial equations, and hence usage of rather complex methods like Gröbner basis. It is also likely that we will get multiple solutions and not a single one. 
Finally, we could plug $\lambda_1$ and $\lambda_2$ into the solution parameterization (see above) to get $\mathbf{q}$. Recovering $\alpha_1$ and $\alpha_2$ will then be very simple:
$$
\alpha_1 = \arccos(\mathbf{q}^1)\\
\alpha_2 = \arccos(\mathbf{q}^3)\\
$$
