How do I solve $\sin{a} + \sin{b} + \sin{c} = 2 \land \cos{a} + \cos{b} + \cos{c} = 2 $? How do I solve $\sin{a} + \sin{b} + \sin{c} = 2 \land \cos{a} + \cos{b} + \cos{c} = 2 $?
I tried the following in Mathematica, but it did not give any solutions. I would appreciate an analytical solution (or the proof that no solutions exist).
Solve[{
  Sin[a] + Sin[b] + Sin[c] == 2,
  Cos[a] + Cos[b] + Cos[c] == 2},
 {a, b, c}]

 A: What it looks to me is that you want to make $(2,2)=\vec{a}+\vec{b}+\vec{c}$ where $\vec{a}, \vec{b}, \vec{c}$ are unit vectors in the Euclidean plane. Here is a way to do it geometrically:
(a) Draw a circle around $O=(0,0)$ with radius $1$ and a circle around $C=(2,2)$ with radius $2$. They will intersect in two points $X,Y$. Take a point $A$ anywhere on the smaller arc $\overset{\frown}{XY}$, and let $\vec{a}=\vec{OA}$. $\vec{a}$ will be a unit vector.
(b) Now make two circles of radius $1$ centered in $A$ and $C$. They will intersect in one or two points: take any of them (call it $B$) and now both $\vec{b}=\vec{AB}$ and $\vec{c}=\vec{BC}$ will be unit vectors, and $\vec{a}+\vec{b}+\vec{c}=\vec{OC}=(2,2)$.
A: This question reminds me of articulated mechanisms (Watt, Peaucelier, Hart, Chebyshev...), for example the one that is represented at the top of this page of a huge compendium of such mechanisms.
It is why I still propose geometrical approaches (sorry...) that are  efficient : they allow to obtain exact solutions (formulas (11)) or, using a different approach, approximate solutions (formulas (5)).
Let us write down for convenience the given equations:
$$\begin{cases}\cos a+\cos b+\cos c&=&2\\ \sin a+\sin b+\sin c&=&2\end{cases}\tag{1}$$
I would like to give a first geometrical representation of (1), with points
$$B=\binom{\cos a}{\sin a}, \ C=\binom{\cos a+\cos b}{\sin a+\sin b}, \ D=\binom{\cos a+\cos b+\cos c=2}{\sin a+\sin b+\sin c=2}\tag{2}$$
in a system of rectangular coordinates (Fig. 1).
One can see positions of $A,B,C,D$ in an ordinary case and (in red) $A,B',C',D$ representing a limit case.

Fig. 1.
Fig. 2 provides a representation of 3 surfaces, now in  $(a,b,c)$ coordinate space:
$$\begin{cases}c&=&\cos^{-1}(2-\cos a - \cos b) & \ (blue) \\
c&=&\sin^{-1}(2-\sin a - \sin b) & \ (purple) \\
c&=&\pm \cos^{-1}(\tfrac{1}{4 \sqrt{2}}(7-2 \cos(a-b))) & \ (red)\end{cases}\tag{3}$$

Fig. 2.
The first and second equations come clearly from (1).
The third equation is issued from the squaring and adding of equations :
$$\begin{cases}2-\cos c&=&\cos a+\cos b\\ 2-\sin c&=&\sin a+\sin b\end{cases}\tag{4}$$
The interest of the third equation is that, apart from its cylindrical representation, it gives the following constraint:
$$\cos(a-b) \ge \frac12 (7-4 \sqrt{2})\approx 0.67157$$
that can be exploited later on.
Remark: In Fig. 2, we have taken a $[0,\pi]^3$ box in order to have a broader view, but in fact solutions to (1) belong to $[0,\pi/2]^3$. Had we taken an even larger scope, we would have seen that the surface represented by the second equation generates a bowl-like surface identical to the first surface, but for the fact that it is a bowl turned upside down.
Fig. 2 shows that these surfaces share a common curve ; otherwise said, the set of solutions $(a,b,c)$ is a one-dimensional loop, approximately circular, in the $(a,b,c)$  coordinate system.

Edit 1: on an experimental basis, I have found that the intersection curve (otherwise said, the general solution to equations (1)) is very close (see Fig. 3) to the 3D curve with equations:
$$\begin{cases}a&=&\dfrac{\pi}{4}+0.245 \cos t - 0.42 \sin t \\
b&=&\dfrac{\pi}{4}+0.245 \cos t + 0.42 \sin t \\
c&=&\dfrac{\pi}{4}-0.49\cos t\end{cases}\tag{5}$$
Otherwise said, an approximate solution of (1) (giving RHSides between $1.99$ and $2.01$ for any value of $t$) is:
$$\begin{pmatrix}a\\b\\c \end{pmatrix}=\begin{pmatrix}\dfrac{\pi}{4}\\ \dfrac{\pi}{4}\\ \dfrac{\pi}{4} \end{pmatrix}+\cos t \underbrace{\begin{pmatrix}+0.245\\+0.245\\-0.49  \end{pmatrix}}_U+\sin t \underbrace{\begin{pmatrix}-0.425  \\ +0.425 \\ 0 \end{pmatrix}}_V\tag{6}$$
With $U,V$ forming an orthogonal basis of plane $(P)$ with equation $a+b+c=0$; formulas (6) describe (up to numerical approximations) the circle with center $(\pi/4,\pi/4,\pi/4)$ and radius $\approx 0.6$ belonging to plane $a+b+c=3 \pi/4$.
There is a slight discrepancy with a pure circle: as we can see on Fig. 3, the resulting curve is partly in front of the two surfaces (the 3 continuous arcs) and partly behind them (the 3 dotted arcs).

Fig. 3.
Edit 2: Following exchanges with @dxiv, I have realized that a complex approach is rather direct.
The issue being to find triples $(a,b,c)$ such that
$$e^{ia}+e^{ib}+e^{ia}=2+2i=2\sqrt{2}e^{i \pi/4}\tag{7}$$
which, by setting
$$a=\pi/4+\alpha, \ b=\pi/4+\beta, \ c=\pi/4+\alpha\tag{8}$$
can be transformed into:
$$\underbrace{e^{i \alpha}+e^{i \beta}}_{2z}+e^{i \gamma}=2\sqrt{2}\tag{9}$$
where $z$ is the midpoint of the line segment joining $e^{i \alpha}$ and $e^{i \beta}$ (represented as point $D$ on Fig. 4) ; as this line segment is orthogonal to the line joining $O$ and $z$, we can write, using (9) and what we just said above:
$$\begin{cases}z&:=&\frac23 \sqrt{2}-\tfrac12 e^{i \gamma}=\cos \phi e^{i \delta}& \ (point D)\\
e^{i \alpha}&:=&e^{i \delta}e^{i \phi}& \ (point A)\\
e^{i \beta}&:=&e^{i \delta}e^{-i \phi}& \ (point B)\end{cases}\tag{10}$$

Fig. 4: The blue point is the barycenter $(\frac23 \sqrt{2}, \ 0)$ of points $A=e^{i \alpha}, \ B=e^{i \beta}, \ C=e^{i \gamma}, \ $. The materialized angle has value $\phi$.
Being given angle $\gamma$, using relationships (8), we are able to deduce the two other angles $\alpha$ and $\beta$:
$$\alpha,\beta=\tan^{-1}\left(\frac{\sin \gamma}{\cos \gamma -2 \sqrt{2} }\right)\pm \cos^{-1}\left(\sqrt{\frac94-\sqrt{2} \cos \gamma}\right)\tag{11}$$

Fig. 5. The curves of $\alpha,\beta$ (in green and red) as functions $\alpha(\gamma),\beta(\gamma)$ of variable $\gamma$ given by (11). One notices the perfect symmetry between $a$ and $b$, with an almost elliptic curve.
(and then adding $\pi/4$ to retrieve values of $a,b,c$ (formulas (8)).
Rather simple calculations show that angles $\alpha, \beta, \gamma$ should be in the range $[-L,L]$ where $L=\cos^{-1}(\tfrac58 \sqrt{2})=27.89°$.
A: The following will consider the slightly more general form:
$$
\begin{align}
\cos{a} + \cos{b} + \cos{c} &= u \tag{1}
\\ \sin{a} + \sin{b} + \sin{c} &= v \tag{2}
\end{align}
$$
Let $\,\sigma = u + i v\,$, $\,\alpha=\cos a + i \sin a\,$ and similar for $\,\beta,\gamma\,$ with $\,|\alpha|=|\beta|=|\gamma|=1\,$, then adding $\,(1) + i\,\cdot (2)\,$ translates to $\,\alpha+\beta+\gamma= \sigma\,$. Taking conjugates and using that $\,\overline\alpha = 1 / \alpha\,$ etc allows solving for $\,\alpha\,$, $\,\beta\,$ in terms of $\,\sigma\,$, $\,\gamma\,$:
$$
\begin{align}
\alpha + \beta &= \sigma - \gamma
\\ \frac{1}{\alpha}+\frac{1}{\beta} &= \overline\sigma-\overline\gamma \;\;\implies\;\; \alpha\beta = \frac{\alpha+\beta}{\overline\sigma-\overline\gamma} = \frac{\sigma - \gamma}{\overline\sigma-\overline\gamma}
\end{align}
$$
With $\displaystyle\,\lambda=\frac{1}{\sigma-\gamma}\,$ it follows that $\,\alpha\,$, $\,\beta\,$ are the roots of the quadratic:
$$
\begin{align}
t^2 - \frac{1}{\lambda} t + \frac{\overline\lambda}{\lambda} = 0 \;\;\;\;\iff\;\;\;\; &\lambda t^2 - t+ \overline \lambda = 0
\\ \;\;\;\;\iff\;\;\;\;  & \alpha,\beta = \frac{1 \pm i\sqrt{4 |\lambda|^2-1}}{2\lambda} \tag{3}
\end{align}
$$
The roots $\,\alpha,\beta\,$ will have modulus $\,1\,$ iff $\,4 |\lambda|^2 \ge 1\,$ $\,\iff |\sigma-\gamma| \le 2\,$. Therefore the system has solutions when at least one $\,\gamma\,$ exists such that $\,|\gamma|=1\,$ and $\,|\sigma-\gamma| \le 2\,$, which is equivalent to $\,|\sigma| \le 1+2=3\,$. In that case, for each such $\,\gamma\,$ there exists a solution $\,\alpha\,$, $\,\beta\,$ given by $\,(3)\,$, which is unique up to a permutation of variables.

[ EDIT ] $\;$ Inspired by Jean Marie's answer, here is a different way to parameterize the solution set.
The condition $\,|\sigma-\gamma| \le 2\,$ was previously established. Moreover, the triangle inequality gives the additional constraints $\,\big||\sigma|-1\big|\le|\sigma-\gamma|\le |\sigma|+1\,$. Then $\,\sigma-\gamma\,$ can be written in the form:
$$
\sigma-\gamma = 2\omega\,\cos\varphi
\;
\begin{cases}
|\omega|=1
\\
\varphi \in [\varphi_{min}, \varphi_{max}] \subseteq [0, \pi/2]
\;
\begin{cases}
\varphi_{min}= \arccos \min\left(\frac{|\sigma|+1}{2}, 1\right)
\\ \varphi_{max}=\arccos \frac{\big||\sigma|-1\big|}{2} \tag{4}
\end{cases}
\end{cases}
$$
For each $\,\varphi \in [\varphi_{min}, \varphi_{max}]\,$ there will exist two $\,\omega\,$ such that $\,|\omega|=1\,$ and $\,|\sigma - 2 \omega \cos \varphi|$ $= |\gamma| = 1\,$, which are the roots of the equation below (it can be shown that the quadratic always has two complex roots $\,\omega_{1,2}\,$ of modulus $\,|\omega_1|=|\omega_2|=1\,$ when $\,\varphi\,$ is in the given range):
$$
|\sigma - 2 \omega \cos \varphi|^2=1 \iff 2 \overline \sigma \cos \varphi \cdot \omega^2 - \left(|\sigma|^2 + 4 \cos^2 \varphi - 1\right)\cdot\omega + 2\sigma\cos\varphi = 0 \tag{5}
$$
With either $\,\sigma-\gamma=2\omega \cos\varphi\,$, $\,\omega \in \{\omega_1, \omega_2\}\,$ it follows that $\,2\lambda = \frac{1}{\omega\cos\varphi}\,$, then $\,(3)\,$ becomes:
$$
\alpha,\beta = \frac{1 \pm i\sqrt{\frac{1}{\cos^2\varphi}-1}}{\frac{1}{\omega\cos\varphi}} = \omega\left(\cos\varphi \pm i \sin\varphi\right)
$$
So, in the end, the solution set is:
$$
\begin{cases}
\begin{align}
\alpha &= \omega(\cos\varphi + i \sin\varphi)
\\ \beta &= \omega(\cos\varphi - i \sin\varphi)
\\ \gamma &= \sigma - 2\omega\cos\varphi \qquad\qquad\qquad \text{where}\;\; \varphi \in [\varphi_{min}, \varphi_{max}] \text{ from (4)}\,, \;\,\omega \text{ from (5)}
\end{align}
\end{cases}
$$
