Solving $2\sin(\omega x)+\sin(\omega y)=0$, $\;2y-4x=c$ for $x$ and $y$. Is there a closed form?

Question

I try to solve the following problem, which consists simply of 2 equations and 2 unknowns.

\begin{align} 2\sin(\omega x)+\sin(\omega y) &= 0 \tag1\\ 2y - 4x &= c \tag2 \end{align} Suppose $\omega$ and $c$ are known variables. I try to solve for the variable $x$ and $y$. I tried to take $y=0.5c+2x$ and plug it into Eq. $(1)$, which gives me \begin{align} 2\sin(\omega x) + \sin(\omega(0.5c+2x))= 0 \tag3\\ 2\sin(\omega x) + \sin(0.5\omega c)\cos(2\omega x) + \sin(2\omega x)\cos(0.5\omega c)= 0 \tag4\\ 2\sin(\omega x) + \sin(0.5\omega c)(1-2\sin^2(\omega x)) + 2\sin(\omega x)\cos(\omega x)\cos(0.5\omega c)= 0 \tag5\\ 2\sin(\omega x)\left[1-\sin(0.5\omega c)\sin(\omega x) + \cos(\omega x)\cos(0.5\omega c)\right]= -\sin(0.5\omega c) \tag6 \end{align}

From here I feel that I am stuck and can't further simplify. I tried already to solve this problem in Wolfram Alpha, MAPLE and in MATLAB but they all show me very complex solutions. I was wondering if this problem has a closed form expression for $x$ and $y$? I appreciate any input, which I can get. Thank you for your time.

Answer 1

To make it simpler, let $c=2a \omega$ which gives $y=2x+a \omega$.

So the first equation becomes $$2 \sin (x \omega )+\sin (a \omega^2+ 2x\omega )=0$$ Expand the second sine to get $$2 \sin (x \omega )+\sin (a \omega^2)\cos(2x\omega )+\cos (a \omega^2)\sin(2x\omega )=0$$

Let $\alpha=\sin (a \omega^2)$, $\beta=\cos (a \omega^2)$ and $t=x \omega$ to get $$2\sin(t)+\alpha \cos(2t)+\beta \sin(2t)=0$$

Let $t=\cos ^{-1}(z)$ and obtain $$\alpha \left(2 z^2-1\right)+2 \sqrt{1-z^2} (\beta z+1)=0$$ which, by squaring, is a quartic in $z$ $$\left(\alpha ^2-4\right)-8 \beta z-4 z^2 \left(\alpha ^2+\beta ^2-1\right)+8 \beta z^3+4 z^4 \left(\alpha ^2+\beta ^2\right)=0$$

Now, using the relations between $\alpha$ and $\beta$, this becomes $$-\left(\beta ^2+3\right)-8 \beta z+8 \beta z^3+4 z^4 =0$$ which is solvable using radicals.

The discriminant is $$\Delta=-110592 \left(\beta ^4-3 \beta ^2+2\right)^2$$ which is always negative. Then the equation has two distinct real roots and two complex conjugate non-real roots.

You must take care that squaring introduces spurious solutions. Then check them.

Edit

Using $t=2\tan^{-1}(w)$, the equation write $$\alpha +4(1+ \beta ) w-6 \alpha w^2+4(1- \beta ) w^3+\alpha w^4=0$$ for which $$\Delta=-110592 \left(1-\beta ^2\right)$$ So, again two distinct real roots and two complex conjugate non-real roots.

Answer 2

If $c=0 $

Plug the second relation into the first..

$$\sin \omega x + \sin 2 \omega x = 0;\quad \sin \omega x = 0 \to \omega x= k \pi $$

$$\cos \omega x = =-1 \to \omega x = (2 j+1) \pi $$

Here we have the real roots/ solutions as shown

$ \pi\text{ double root} , 2 \pi, 3 \pi\text{double root} , 4 \pi, 5 \pi\text{ double root} ...$ etc. $y$ roots are double those of $x$ roots.

When $ c\ne 0 $ we have in addition complex conjugate roots which can be found by asymptotic expansions... as well as by numerical iteration with Newton Raphson...