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. 

Suppose you have instead
$$
e^{ix}+e^{iy} = a\\
e^{2ix}+e^{2iy} = b
$$
then
$$
e^{2ix}+e^{2iy}+2e^{ix}e^{iy} = a^2\Rightarrow 2e^{ix}e^{iy} =a^2-b
$$
and now
$$
e^{ix}+e^{iy} = a\\
e^{ix}e^{iy} =\frac{a^2-b}{2}
$$
etc.
NOTE
Another way is to follow the trig. identities
$$
\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) = \frac a2\\
\sin(x+y)\cos(x-y) = \frac a2
$$
now calling 
$$
\frac{x+y}{2} = u\\
\frac{x-y}{2} = v
$$
$$
\sin^2u\cos^2v = (1+\cos(2u))(1-\cos(2v)) = a^2
$$
or
$$
1+\cos(2v)-\cos(2u)-\cos(2u)\cos(2v) = a^2\\
\sin(2u)\cos(2v) = \frac a2
$$
and
$$
\sin(2u) = \frac a2\left(\frac{\cos(2u)-1}{1-a^2+\cos(2u)}\right)
$$
and now calling $\cos(2u) = z$ we have
$$
\left(\frac 2a\right)^2\left(\frac{z-1}{1-a^2+z}\right)^2=1-z^2
$$
This quartic should be solved to obtain the solutions. Attached a plot showing the intersections between the curves
$$
\sin x + \sin y = \frac 15 \;\;\mbox{in red}\\
\sin(2x)+\sin(2y) = \frac 15\;\;\mbox{in blue}
$$

