I've been struggling to solve equations in two variables containing constant symbolic parameters. Take for example the function in two variables defined by:

F[x_,y_]:=Sqrt[2 t1^2 + t3^2 + 4 t1 t3 Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 2 t1^2 Cos[Sqrt[3] y]]

My goal is to solve $$F(x,y)=0$$ so I want to find the coordinates of the points in the 2D xy plane where the function is 0. These coordinates will be given by an expression containing the parameters $t_{1}$, $t_{3}$. I can find some solutions by guessing, for example: $$(x,y)=(0,\frac{2}{\sqrt{3}}\arccos{(-\frac{t_{3}}{2t_{1}})}) $$

or $$(x,y)=(0,-\frac{2}{\sqrt{3}}\arccos{(-\frac{t_{3}}{2t_{1}})}) $$

I would like to find these solutions (and the others) in Mathematica,so I've tried:

Solve[F[x, y] == 0, {x, y}]


Reduce[F[x, y] == 0, {x, y}]

However my problem is that Mathematica gives me an expression where the solution for x contains y or viceversa, but I would like the solutions separate (so the solution for x shall not contain y and viceversa, as in the two I wrote above). Also, sometimes the parameters $t_{1}$,$t_{3}$ are treated as variables by Mathematica, and a solution is provided also for them, which I don't want.

This is a problem I had in general for these kind of equations where I solve for 2 variables and I also have additional constant parameters. So how can I get the solution for x and y (without them been coupled) in terms of the constant parameters?

  • $\begingroup$ Have a look at Solve. $\endgroup$ Commented Aug 26, 2022 at 13:08
  • $\begingroup$ I've used that, but it gives me the problem I explained. $\endgroup$
    – Mathew
    Commented Aug 26, 2022 at 14:04
  • 1
    $\begingroup$ There are quit a lot of solutions as Reduce shows: Reduce[F[x, y] == 0, {x, y}, Reals] $\endgroup$ Commented Aug 26, 2022 at 15:04
  • $\begingroup$ With Solve use Solve[F[x, y] == 0, {x, y}, Reals, Method -> Reduce] $\endgroup$
    – Bob Hanlon
    Commented Aug 26, 2022 at 15:38

1 Answer 1


Solve for Reals with Method->Reduce like @BobHanlon proposes.

Plot solutions for x and y in dependence of t1 and t3, or get values at definite t1,t3.

F[x_, y_] = 
  Sqrt[2 t1^2 + t3^2 + 4 t1 t3 Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 
    2 t1^2 Cos[Sqrt[3] y]]

sol3[c1_, c2_, t1_, t3_] = 
 Solve[F[x, y] == 0, {x, y}, Reals, Method -> Reduce] /. {C[1] -> c1, 
   C[2] -> c2}

Plot3D[Evaluate[y /. sol3[0, 0, t1, t3] // Simplify // Union], {t1, 0,
   1}, {t3, 0, 1}, PlotStyle -> {Red, Green, Blue, Magenta}, 
 PlotRange -> All]

{x, y} /. sol3[0, 0, 1/3, 1/2] // TableForm

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