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. 

First,$$
\begin{cases}
\sin x + \sin y = \dfrac{1}{5}\\
\sin 2x + \sin 2y = \dfrac{1}{5}
\end{cases} \Longrightarrow \begin{cases}
\sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} = \dfrac{1}{10} & (1)\\
\sin(x + y) \cos(x - y) = \dfrac{1}{10} & (2)
\end{cases}.
$$
Denote $u = \cos \dfrac{x - y}{2}$, $v = \sin \dfrac{x + y}{2}$, then (1) and (2) imply$$
\begin{cases}
uv = \dfrac{1}{10}\\
(2u^2 - 1) v \sqrt{1 - v^2} = \pm \dfrac{1}{20}
\end{cases} \Longrightarrow \begin{cases}
uv = \dfrac{1}{10} & (3)\\
(2u^2 - 1)^2 v^2 (1 - v^2) = \dfrac{1}{400} & (4)
\end{cases}.
$$
Since $v = \dfrac{1}{10u}$ by (3), eliminating $v$ from (4) yields$$
(2u^2 - 1)^2\left( 4u^2 - \frac{1}{25} \right) = u^4\\
\Longrightarrow 16 (u^2)^3 - \left( \frac{4}{25} + 17 \right) (u^2)^2 + \left( \frac{4}{25} + 4 \right) u^2 - \frac{1}{25} = 0.
$$
Note that $-π < y \leqslant x \leqslant π$ implies $l$ Now, there are six solutions to the last equation (see WA), i.e. $u_1, \cdots, u_6$. For each $u_k$, there is $v_k = \dfrac{1}{10u_k}$. Note that $\cos \dfrac{x + y}{2} = \pm \sqrt{1 - v^2}$. If $(2u_k^2 - 1) v_k > 0$, then$$
\frac{1}{20} = (2u_k^2 - 1) v_k \cos \frac{x + y}{2} \Longrightarrow \cos \frac{x + y}{2} = \sqrt{\smash[b]{1 - v_k^2}}.
$$
Otheriwise $(2u_k^2 - 1) v_k < 0$, then$$
\frac{1}{20} = (2u_k^2 - 1) v_k \cos \frac{x + y}{2} \Longrightarrow \cos \frac{x + y}{2} = -\sqrt{\smash[b]{1 - v_k^2}}.
$$
Thus for each $k$,$$
\begin{cases}
\cos \dfrac{x - y}{2} = u_k,\ \sin \dfrac{x - y}{2} = \pm \sqrt{\smash[b]{1 - u_k^2}}\\
\sin \dfrac{x + y}{2} = v_k,\ \cos \dfrac{x + y}{2} = ε_k \sqrt{\smash[b]{1 - v_k^2}}
\end{cases},
$$
where $ε_k \in \{\pm 1\}$ is determined as above. Since$$
\begin{cases}
\sin x = \sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} + \cos \dfrac{x + y}{2} \sin \dfrac{x - y}{2}\\
\cos x = \cos \dfrac{x + y}{2} \cos \dfrac{x - y}{2} - \sin \dfrac{x + y}{2} \sin \dfrac{x - y}{2}\\
\sin y = \sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} - \cos \dfrac{x + y}{2} \sin \dfrac{x + y}{2}\\
\cos y = \cos \dfrac{x + y}{2} \cos \dfrac{x - y}{2} + \sin \dfrac{x + y}{2} \sin \dfrac{x - y}{2}
\end{cases},
$$
there are twelve possible solutions $(\sin x, \cos x, \sin y, \cos y)$. After computing them numerically, it turns out that there are six distinct solutions.
