Solve the following system of equations $$\begin{cases} \sin(a)+2\sin(a+b+c)=0 \\ \sin(b)+3\sin(a+b+c)=0 \\ \sin(c)+4\sin(a+b+c)=0 \end{cases}$$
I added all the three equations and got $$\frac{\sin(a)+\sin(b)+\sin(c)}{9}+\sin(a+b+c)=0.$$ Then, I have very vague view on how to continue is using something like Jensen's inequality... Any help is appreciated.
I am using that $\sin a = A \Rightarrow a = \sin^{-1} A$, which is true if they are in $[- \frac{\pi}{2}, \frac{\pi}{2}]$.
Let $ \sin a = A, \sin b = B, \sin c = C$.
You have shown that $ \sin (a+b+c) = -\frac{ A + B + C } { 9 } $.
Substituting back in, we get
$7A -2B - 2C = 0 $
$-3A + 6 B - 3C = 0$
$ -4 A - 4 B + 5 C = 0 $
Solutions to this system are of the form $ A:B:C = 2:3:4$.
Let $ A = 2n, B = 3n, C = 4n$ where $ -\frac{1}{4} \leq n \leq \frac{1}{4}$.
We then need to check against the condition of
$-9\sin (a+b+c ) = A+B+C = 9n $, which gives us
$\sin^{-1} (2n) + \sin^{-1} (3n) + \sin^{-1} (4n) = - \sin^{-1}n $ or $\pi + \sin^{-1} (n) $ or $ 2\pi - \sin^{-1} (n)$.
For $n \in ( 0, \frac{1}{4})$, we can show that the LHS is $ \in (0, 3 )$, hence is never equal to the RHS. A similar statement holds for $ n \in ( - \frac{1}{4} , 0)$.
So, the only solution is $ n = 0 \Rightarrow A=B=C = 0 $.
To extend outside of the range, we need to consider $ a = \pi - \sin^{-1} A$, which makes it seem like solutions exist. To be continued (?) ...
Expand out $\sin(a+b+c)$, take $\sin(a)=s_a$, $\cos(a)=c_a$, etc., adjoin the relations $ s_a^2 + c_a^2 - 1 = 0$ etc, and we can solve using Groebner basis methods. All solutions have $\sin(a) = \sin(b) = \sin(c) = 0$, i.e. $a,b,c$ are integer multiples of $\pi$.
