<s>[INST] Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$ To find the general solution $(x,y)$ satisfying the system of equations
\begin{align} 5(\sin x + \sin y) &= 1 \\ 5(\sin 2x + \sin 2y) &= 1
\end{align}
I applied $\sin C + \sin D$ rule and then divided these two equations, then I am stuck at
$$\cos\frac{x-y}{2} = 2\cos \frac{x+y}{2}\cos(x-y).$$
I do not know what to do further. 
 [/INST] Starting from Alex Francisco's answer
$$16 (u^2)^3 - \left( \frac{4}{25} + 17 \right) (u^2)^2 + \left( \frac{4}{25} + 4 \right) u^2 - \frac{1}{25} = 0$$ let $x=u^2$ to get the cubic
$$400 x^3-429 x^2+104 x-1=0$$ and use the trigonometric method for three real roots. This would give the nasty
$$x_k=\frac{143}{400}+\frac{7}{200} \sqrt{\frac{403}{3}} \cos \left(\frac{2 \pi  }{3}k-\frac{1}{3} \cos
   ^{-1}\left(\frac{89421
   \sqrt{{3}}}{138229\sqrt{403}}\right)\right)\qquad (k=1,2,3)$$ which are all positive. Then six solutions for $u$ and $v$.
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