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I'm stuck on this and chagrined I'm even asking, but this:

NSolve[{-12 + 16 Cos[t]^2 == 4 - 2 s^3, 
  4 - 2 t^3 == s^3, -π <= s < π, -π <= t < π}, {s, t}]

gives "NSolve::nsmet: This system cannot be solved with the methods available to NSolve", or Solve for that matter. Thank you

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  • 1
    $\begingroup$ From the documentation: "NSolve deals primarily with linear and polynomial equations". This is more of a problem for FindRoot. $\endgroup$ Commented Oct 21, 2020 at 15:00

5 Answers 5

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You could solve it as a minimization:

NMinimize[{(-12 + 16 Cos[t]^2 - (4 - 2 s^3))^2 + ( 
    4 - 2 t^3 - s^3)^2, -π <= s < π, -π <= t < π}, {s, t}]
(* result: {3.16777*10^-30, {s -> 1.49116, t -> 0.699433}} *)

system = {-12 + 16 Cos[t]^2 == 4 - 2 s^3, 4 - 2 t^3 == s^3, -π <= s < π, -π <= t < π};
system /. {s -> 1.4911575481418087`, t -> 0.6994333479233502`}

(* result: {True, True, True, True} *)
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Clear["Global`*"]

eqns = {-12 + 16 Cos[t]^2 == 4 - 2 s^3, 4 - 2 t^3 == s^3};

The exact solution for t is a Root expression

solt = Solve[{Eliminate[eqns, s], -π <= t < π}, t][[1]]

enter image description here

solt // N

(* {t -> 0.699433} *)

The exact solution for s is similarly

sols = Solve[{eqns[[2]] /. solt, -π <= s < π}, s][[1]]

enter image description here

sols // N

(* {s -> 1.49116} *)

(eqns /. solt /. sols) // N

(* {True, True} *)
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  • $\begingroup$ Strangely, your Solve[{t^3 == 2 Cos[2 t], -π < t < π}, t] works but the equivalent Solve[t^3 == 2 Cos[2 t], t] does not work. Your solution is better than the FindRoot application I proposed. $\endgroup$
    – Roman
    Commented Oct 21, 2020 at 15:31
  • 1
    $\begingroup$ @Roman - in this case, the constraint is equivalent to specifying the domain as real, i.e., Solve[t^3 == 2 Cos[2 t], t, Reals] $\endgroup$
    – Bob Hanlon
    Commented Oct 21, 2020 at 16:23
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Substitute the second equation into the first:

FullSimplify[-12 + 16 Cos[t]^2 == 4 - 2 s^3 /. s^3 -> 4 - 2 t^3]
(*    t^3 == 2 Cos[2 t]    *)

By inspection we see that there is only one solution:

Plot[{t^3, 2 Cos[2 t]}, {t, -2, 2}, PlotRange -> {-3, 3}]

enter image description here

Find the solution and compute $\{s,t\}$:

{(4 - 2 t^3)^(1/3), t} /. FindRoot[t^3 == 2 Cos[2 t], {t, 1}]
(*    {1.49116, 0.699433}    *)
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NSolve can solve the system as complex problem:

csol = NSolve[{-12 + 16 Cos[t]^2 == 4 - 2 s^3, 
   4 - 2 t^3 == s^3, -π <= Re@s < π, -π <= Re@t < π,
    -1/10 < Im@s < 1/10, -1/10 < Im@t < 1/10}, {s, t}]
rsol = Chop[csol]
(*
  {{s -> 1.49116 + 0. I, t -> 0.699433 + 0. I}}
  {{s -> 1.49116, t -> 0.699433}}
*)

Update

Solve also works if you avoid the singularity at s == 0:

Solve[{-12 + 16 Cos[t]^2 == 4 - 2 s^3, 4 - 2 t^3 == s^3,
  1 < s < 2, -π <= t < π}, {t, s}, Reals]
(*
{{t -> Root[{2 - 4 Cos[#1]^2 + #1^3 &, 0.69943334792335012145}], 
  s -> Root[-4 + 2 Root[{2 - 4 Cos[#1]^2 + #1^3 &, 
         0.69943334792335012145}]^3 + #1^3 &, 1]}}
*)

Also if it is "factored" out (as a product $(s,t)\times (u)$):

Solve[{-12 + 16 Cos[t]^2 == 4 - 2 u, 4 - 2 t^3 == u, u == s^3,
  -π <= s < π, -π <= t < π}, {t, s, u}, Reals]
(*
{{t -> Root[{2 - 4 Cos[#1]^2 + #1^3 &, 0.69943334792335012145}], 
  s -> Root[-8 + 8 Cos[Root[{2 - 4 Cos[#1]^2 + #1^3 &, 
          0.69943334792335012145}]]^2 + #1^3 &, 1], 
  u -> 8 - 8 Cos[Root[{2 - 4 Cos[#1]^2 + #1^3 &, 
          0.69943334792335012145}]]^2}}
*)
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Use ContourPlot and Graphics`MeshFindIntersections.

And then use FindRoot again.

Clear["`*"];
Clear["`*"];
curves = ContourPlot[{-12 + 16 Cos[t]^2 == 4 - 2 s^3, 
   4 - 2 t^3 == s^3}, {s, -π, π}, {t, -π, π}, 
  PlotPoints -> 200, 
  ContourStyle -> {Directive[Red, Thickness[Tiny]], 
    Directive[Blue, Thickness[Tiny]]}, ImageSize -> Full]
Graphics`Mesh`FindIntersections[curves[[All, 1]]]
(* {{1.49114,0.699426}} *)

FindRoot[{-12 + 16 Cos[t]^2 == 4 - 2 s^3, 4 - 2 t^3 == s^3}, {s, 
  1.49114}, {t, 0.699426}]
(* {s -> 1.49116, t -> 0.699433} *)

enter image description here

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