Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$

Question

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.

Answer 1

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.

Answer 2

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

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

Answer 3

Easy to see that \begin{cases} \sin y = \dfrac15-\sin x\\ 2\sin y\cos y = \dfrac15 - \sin2x \end{cases} \begin{cases} \sin y = \dfrac15-\sin x\\[4pt] \cos y = \dfrac12\dfrac{1-5\sin 2x}{1-5\sin x}.\\ \end{cases} Then $$\dfrac1{25}\left(1-5\sin x\right)^2 + \dfrac14\left(\dfrac{1-5\sin 2x}{1-5\sin x}\right)^2=1,$$ $$4(1-5\sin x)^4+25(1-5\sin 2x)^2 = 100(1-5\sin x)^2,$$ $$4\left(\sin^2\dfrac x2-10\sin \dfrac x2\cos \dfrac x2 + \cos^2\dfrac x2\right)^4+25\left(\left(\sin^2\dfrac x2 + \cos^2\dfrac x2\right)^2-20\sin \dfrac x2\cos\dfrac x2\left(\cos^2 \dfrac x2 - \sin^2 \dfrac x2\right)\right)^2 = 100\left(\sin^2\dfrac x2-10\sin \dfrac x2\cos \dfrac x2 + \cos^2\dfrac x2\right)^2\left(\sin^2\dfrac x2 + \cos^2\dfrac x2\right)^2,$$ and the subsitution $$t=\tan\dfrac x2$$ leads to the algebraic equation $$4\left(t^2+1-10t\right)^4+25\left((t^2+1)^2-20t(1-t^2)\right)^2 = 100\left(t^2+1-10t\right)^2\left(t^2+1\right)^2,$$ or $$(t^2+1)(71t^6-2840t^5-2187t^4+12320t^3-2187t^2-840t+71)= 0,$$ with the real roots $$t\in\{-2.475108, -0.227373, 0.0758504, 0.356347, 1.617116, 40.453167\}$$ (see also Wolfram Alpha).

This means that \begin{cases} x_i=2\pi m+2\arctan t_i,\quad m\in\mathbb N\\[4pt] \sin x_i = \dfrac {2t_i}{1+t_i^2}\\[4pt] \sin 2x_i = \dfrac{4t_i(1-t_i^2)}{(1+t_i^2)^2}.\tag1 \end{cases} On the other hand, $$\tan\dfrac{y}2 = \dfrac{\sin y}{1+\cos y} = \dfrac{\dfrac15-\dfrac{2t}{1+t^2}}{1+\dfrac12\dfrac{1-5\dfrac{4t(1-t^2)}{(1+t^2)^2}}{1-5\dfrac{2t}{1+t^2}}} = \dfrac25\dfrac{(1+t^2-10t)^2}{3(1+t^2)^2-40t},$$ $$\tan\dfrac{y_i}2 = \dfrac25\dfrac{(1+t_i^2-10t_i)^2}{3(1+t_i^2)^2-40t_i},$$ with the values $$\tan\dfrac{y_i}2 = \{1.617116, 0.356347, 40.653167, -0.227373, -2.475108, 0.0758504\}$$ (see also Wolfram Alpha).

These allow to calculate the solutions $$\binom{x}{y} \in \left\{\binom{-2.373654}{2.033937}, \binom{-0.447144}{0.684637}, \binom {0.151411}{3.092406}, \binom{0.684637}{-0.447144}, \binom{2.033937}{-2.373654}, \binom{3.092406}{0.151411}\right\}+\binom{2\pi m}{2\pi n},\quad m\in\mathbb N,\quad n \in\mathbb N.$$ Substitution in the original system confirms the correctness of the solutions obtained.

Answer 4

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.

Answer 5

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 $v$.