How to obtain numerical solutions of a two-variable equation?

Question

I have a two-variable function:

f[x_, y_] := (4 y^2 - 6) (Cos[y π] Cos[x] - Cos[2 y π] Sin[x])

How can I find all the numerical solutions for $x$ and $y$ such that $f(x,y)=0$? The domains are $0<x<\pi$ and $0<y<5$.

More precisely, I want to ask Mathematica to consider $x$ with the step $\frac{\pi }{5}$ and provide the values of $y$ for them.

Answer 1

There are infinitely many (continuum) solutions and basically one cannot list them all e.g. because of the Cantor theorem. This is the reason why numerical approach is unsatisfactory. However we can make use of FindRoot to find a finite numerical subset of the solution space.

Numerical solutions

First we set up a net of values of x and starting point of y (we denote them by k in Table)

nsol = Flatten[ Table[{x, y}/.FindRoot[(4y^2 - 6)(Cos[π y]Cos[x] - Cos[2π y] Sin[x]) == 0,
                                          {y, k}], 
                       {x, π/5, 4 π/5, π/5}, {k, 1/3, 14/3, 1/3}], 
                1] // DeleteDuplicates;

Show[ ContourPlot[(4 y^2 - 6) (Cos[π y] Cos[x] - Cos[2π y] Sin[x]) == 0,
                  {x, -π/2, 2π}, {y, -0.1, 5.1}, 
                  PlotPoints -> 50, MaxRecursion -> 3, 
                  Epilog -> {Red, PointSize[0.023], Point[nsol]}], 
      RegionPlot[ Not[0 < x < π && 0 < y < 5], {x, -π/2, 2π}, {y, -0.1, 5.1}]]

Numerical solutions are contained in nsol, we write only a few and hiding 38 of them:

Short[N @ nsol, 3]

{{0.628319, -0.645834}, {0.628319, 0.645834}, <<38>>, {2.51327, 4.35417},
 {2.51327, 5.64583}}

One can see that the first point in nsol doesn't belong to the domain of our interest. We can get rid of such points setting a different net of starting points in FindRoot or using another tools of the system, however this approach might be more involved for more sophisticated transcendental equations. An example of more detailed discussion how to deal with FindRoot can be found at e.g. First positive root.

Exact (symbolic) solution

More interesting is finding an exact descripton of the blue curves which represent symbolic solutions. The form of our equation $f(x,y)=0$ can be factorized into two algebraic expressions and the roots of the original equation is the set-theoretic sum of solutions of two equations: 4 y^2 - 6 == 0 and (Cos[π y] Cos[x] - Cos[2π y] Sin[x]) == 0.

The straight line is the solution of this equation (for x in the range 0 < x < π)

Solve[(4 y^2 - 6) == 0 && 0 < y < 5, y]

 {{y -> Sqrt[3/2]}}

While another equation gives us those periodic curves, moreover it appears that one can solve equation for x obtaining it as a function of y.

Reduce[(Cos[π y] Cos[x] - Cos[2π y] Sin[x]) == 0 && 0 < x < π, x, Reals]

  (Cos[π y] != 0 && Cos[π y] != 0 && 
x == 2 ArcTan[ Sec[π y] (-Cos[2π y] + Cos[π y] Sqrt[(Cos[π y]^2 + 
         Cos[2 π y]^2) Sec[π y]^2])]) ||
((-π + x)/(2 π) \[NotElement] Integers && Cos[π y] == 0 && 
 Cos[2 π y] == 0 && 0 < x < π)

This expression could be further simplified however the main point is the parametrization of curves with

FullSimplify[ x == 2 ArcTan[ Sec[π y] (-Cos[2π y] + Cos[π y] Sqrt[(Cos[π y]^2       
                             + Cos[2π y]^2) Sec[π y]^2])], 
              0 < x < π && 0 < y < 5] // TraditionalForm

For some reasons the standard approach with Reduce on the original equation takes a long time returning quite an involved expression even though the equation is not hard to solve when we deal with different factors of the original function. Finially we plot the graph of the above symbolic solutions with red points denoting numerical ones, yet another time we exploit ContourPlot

Show[
  ContourPlot[{x == -2ArcTan[ Cos[2π y]Sec[π y] - Sqrt[1 + Cos[2π y]^2 Sec[π y]^2]], 
               y == Sqrt[3/2]}, {x, -π/2, 2π}, {y, -0.3, 5.3}, 
              PlotPoints -> 50, MaxRecursion -> 3, 
              ContourStyle -> {{Thick, Darker@Cyan}, Thick, Magenta}, 
              Epilog -> {Red, PointSize[0.016], Point[nsol]}], 
  RegionPlot[ Not[0 < x < π && 0 < y < 5], {x, -π/2, 2π}, {y, -0.3, 5.3}]]

Answer 2

Reduce also seems to work on the original f, though the answer it gives is somewhat long and involves a collection of Tan and ArcTan.

f[x_, y_] := (4 y^2 - 6) (Cos[y \[Pi]] Cos[x] - Cos[2 y \[Pi]] Sin[x]);
Reduce[f[x, y] == 0, {x, y}]