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. 
 A: 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
$$
\sin x + \sin y = \frac 15 \;\;\mbox{in red}\\
\sin(2x)+\sin(2y) = \frac 15\;\;\mbox{in blue}
$$

A: 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.
A: 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 $v$.
A: Perhaps a general method to approach these kinds of problems is of interest to you. Let us define
 $\, X := e^{ix}, \, Y := e^{iy}. \,$ Use these new variables to express
the first equation as a rational function
 $\, 1 - 5(\sin(x) + \sin(y)) =  (2 X Y + 5 i (X Y - 1) (X + Y)) / (2 X Y). \,$
The second equation yields a similar result with $\, X,Y \,$ replaced by $\, X^2,Y^2. \,$ We now want to find when the numerators are both $\,0.\,$
The resultant of the two numerator polynomials, with respect to Y, is the polynomial
 $$ 100 X^5 
(250 i X^6 + (150-125i) X^5 - 290i X^4 - 229 X^3 + 290i X^2 + (150 + 125i) X -250i). $$
Excluding the $\, X^5 \,$ factor, this has $6$ complex roots for $\,X.\,$ Since $\,x\,$ is real, $\, |X| = 1.$ For each root $\,X\,$ there are two values of $\,Y\,$ each differing by the sign of the real part and exactly one of the two is also one of the $6$ roots for $\,X\,$ which makes sense since the two original equations are symmetric with respect to $\,x\,$ and $\,y.\,$ This means that there are $3$ distinct $\,x,y\,$ pairs of solutions.
