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How to further simplify and combine solution sets of trigonometric equations?

Bill Blair

Posted 24 days ago

When solving the following trigonometric equation, the solution set obtained is: Solve[{5 (Sin[5 x + \[Phi]] - Sin[x]) == 0, 0 <= \[Phi] <= 2 \[Pi]}, x, Reals] How to further simplify and combine solution sets of trigonometric equations to obtain the following solution form? {{x -> ConditionalExpression[ 1/6 (\[Pi] - \[Phi] + 2 \[Pi] ConditionalExpression[1, \[Placeholder]]), ConditionalExpression[1, \[Placeholder]] \[Element] Integers && 0 <= \[Phi] <= 2 \[Pi]]}, {x -> ConditionalExpression[ 1/4 (-\[Phi] + 2 \[Pi] ConditionalExpression[1, \[Placeholder]]), ConditionalExpression[1, \[Placeholder]] \[Element] Integers && 0 <= \[Phi] <= 2 \[Pi]]}}

When solving the following trigonometric equation, the solution set obtained is:

Solve[{5 (Sin[5 x + \[Phi]] - Sin[x]) == 0, 
  0 <= \[Phi] <= 2 \[Pi]}, x, Reals]

enter image description here

How to further simplify and combine solution sets of trigonometric equations to obtain the following solution form?

{{x -> ConditionalExpression[
    1/6 (\[Pi] - \[Phi] + 
       2 \[Pi] ConditionalExpression[1, \[Placeholder]]), 
    ConditionalExpression[1, \[Placeholder]] \[Element] Integers && 
     0 <= \[Phi] <= 2 \[Pi]]}, {x -> 
   ConditionalExpression[
    1/4 (-\[Phi] + 2 \[Pi] ConditionalExpression[1, \[Placeholder]]), 
    ConditionalExpression[1, \[Placeholder]] \[Element] Integers && 
     0 <= \[Phi] <= 2 \[Pi]]}}

enter image description here

POSTED BY: Bill Blair

3 Replies

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Michael Rogers

Michael Rogers, Emory University

Posted 20 days ago

I don't know a good way either, but maybe this can give you ideas: Replace[FactorList[5 (Sin[5 x + \[Phi]] - Sin[x]), Trig -> True], {f_, _} :> Solve[TrigReduce[f^2] == 0 /. 1 + Cos[q_] :> 1 - Cos[q + Hold[Pi]], x, Reals], 1] // ReleaseHold // Apply@Join (* {{x -> ConditionalExpression[-(\[Phi]/4) + (PiC[1])/2, Element[C[1], Integers]]}, {x -> ConditionalExpression[-(Pi/6) - \[Phi]/6 + (PiC[1])/3, Element[C[1], Integers]]}} ) It's difficult to combine sequences in the original result, since the instances of `C[1]` are really independent variables. Then there are oddities like this: Solve[{Cos[x] == 1}, x, Reals] ( {{x -> ConditionalExpression[2PiC[1], Element[C[1], Integers]]}} ) Solve[{Cos[x] == -1}, x, Reals] ( {{x -> ConditionalExpression[-Pi + 2PiC[1], Element[C[1], Integers]]}, {x -> ConditionalExpression[Pi + 2PiC[1], Element[C[1], Integers]]}} *) The `Hold[Pi]` in my code gets around the problem with the last example.

I don't know a good way either, but maybe this can give you ideas:

Replace[FactorList[5 (Sin[5 x + \[Phi]] - Sin[x]), Trig -> True], {f_, _} :> 
    Solve[TrigReduce[f^2] == 0 /. 1 + Cos[q_] :> 1 - Cos[q + Hold[Pi]], x, 
     Reals], 1] // ReleaseHold // Apply@Join
(*
{{x -> ConditionalExpression[-(\[Phi]/4) + (Pi*C[1])/2, 
     Element[C[1], Integers]]}, 
  {x -> ConditionalExpression[-(Pi/6) - \[Phi]/6 + (Pi*C[1])/3, 
     Element[C[1], Integers]]}}
*)

It's difficult to combine sequences in the original result, since the instances of C[1] are really independent variables. Then there are oddities like this:

Solve[{Cos[x] == 1}, x, Reals]
(*
{{x -> ConditionalExpression[2*Pi*C[1], Element[C[1], Integers]]}}
*)

Solve[{Cos[x] == -1}, x, Reals]
(* 
{{x -> ConditionalExpression[-Pi + 2*Pi*C[1], 
     Element[C[1], Integers]]}, 
  {x -> ConditionalExpression[Pi + 2*Pi*C[1], 
     Element[C[1], Integers]]}}
*)

The Hold[Pi] in my code gets around the problem with the last example.

POSTED BY: Michael Rogers

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 21 days ago

A simpler situation where this problem arises is for Solve[Sin[x] == 0, x] where Mathematica lists the odd and the even multiples of `Pi` separately. I don't know how to merge them. Somehow, I would expect the following to give `True`, but it does not: first = Or[Element[x/2, Integers], Element[(x - 1)/2, Integers]]; second = Element[x, Integers]; Reduce[Equivalent[first, second], x, Reals]

POSTED BY: Gianluca Gorni

Bill Blair

Posted 21 days ago

Is there no solution to this problem?

POSTED BY: Bill Blair

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