RE: Solve fails
- To: mathgroup at smc.vnet.net
- Subject: [mg33337] RE: [mg33326] Solve fails
- From: "Higinio Ramos" <higra at usal.es>
- Date: Sat, 16 Mar 2002 01:40:16 -0500 (EST)
- References: <200203150051.TAA15038@smc.vnet.net>
- Reply-to: "Higinio Ramos" <higra at usal.es>
- Sender: owner-wri-mathgroup at wolfram.com
Your equation may be expressed in the form
x(Sin[x] - Sin[x^3])==0
and factorizing the lhs
Factor[x(Sin[x]-Sin[x^3]),Trig\[Rule]True]
\!\(2\ x\ Cos[x\/2 + x\^3\/2]\ Sin[x\/2 - x\^3\/2]\).
So, we finally have
\!\(2\ x\ Cos[x\/2 + x\^3\/2]\ Sin[x\/2 - x\^3\/2] == 0\)
where we can obtain the roots.
If you put
\!\(Solve[2\ x\ Cos[x\/2 + x\^3\/2]\ Sin[x\/2 - x\^3\/2] == 0, x]\)
you obtain some of the roots, incluiding complex roots.
Higinio
----- Original Message -----
From: Richard Fateman <fateman at cs.berkeley.edu>
To: mathgroup at smc.vnet.net
Subject: [mg33337] [mg33326] Solve fails
>
> Solve[x*Sin[x]==x*Sin[x^3],x] gives {{}}
>
> but
> Solve[Sin[x]==Sin[x^3],x] returns unchanged
>
> each gives a warning.
>
> There are very many solutions, among them x=0, x=+-1,
> x=+-2.0253909752730426 approximately
>
>
>
- References:
- Solve fails
- From: Richard Fateman <fateman@cs.berkeley.edu>
- Solve fails