Discussion:
Equation system with trigonometric function
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Stefano Milani
2017-07-24 19:15:29 UTC
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Dear all,

I would like to solve this system of equation:

(T) Ky*x3+T0
(phi) Kx*x2+u
(eq1) (-sin(Kx*x2+u-%pi/4)*(-Ky*x3-T0)-bx*x2)/M=0
(eq2) (cos(Kx*x2+u-%pi/4)*(Ky*x3+T0)-by*x3-M*g)/M=0
(eq3) (cos(Kx*x2+u-%pi/4)*(Ky*x3+T0)-by*x3-M*g)/M=0


Where the required term are x2,x3,u
Which command can I use to obtain all solutions?

Best regards


Stefano Milani




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Stavros Macrakis (Σταῦρος Μακράκης)
2017-07-24 19:43:16 UTC
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Not sure I understand your problem.

Do you mean that Ky*x3+T0=0 and Kx*x2+u=0?

What is the purpose of the denominator (/M) in eqN? After all, X/M = 0 is
equivalent to X=0 for all M, except M=0, which is invalid.

eq2 is identical to eq3, so eq3 is redundant.

Please clarify.

-s
Post by Stefano Milani
Dear all,
(T) Ky*x3+T0
(phi) Kx*x2+u
(eq1) (-sin(Kx*x2+u-%pi/4)*(-Ky*x3-T0)-bx*x2)/M=0
(eq2) (cos(Kx*x2+u-%pi/4)*(Ky*x3+T0)-by*x3-M*g)/M=0
(eq3) (cos(Kx*x2+u-%pi/4)*(Ky*x3+T0)-by*x3-M*g)/M=0
Where the required term are x2,x3,u
Which command can I use to obtain *all* solutions?
Best regards
Stefano Milani
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Stefano Milani
2017-07-24 19:54:56 UTC
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I’m sorry,

eq3 is the following:

(eq3) (l*sin(Kx*x2+u)*(Ky*x3+T0))=0

M is a mass, so is positive define (and in the equation is not necessary).
The first two equation (T and phi) are not equation but variable definition used in three following system of three equations.

T = T0+Ky*x3;
phi= Kx*x2+u;

Hope to clarify.

Best regards

Stefano Milani


Da: Stavros Macrakis (ΣταῊρος Μακράκης)
Inviato: lunedì 24 luglio 2017 21:43
A: Stefano Milani
Cc: maxima-***@lists.sourceforge.net
Oggetto: Re: [Maxima-discuss] Equation system with trigonometric function

Not sure I understand your problem.

Do you mean that Ky*x3+T0=0 and Kx*x2+u=0?

What is the purpose of the denominator (/M) in eqN?  After all, X/M = 0 is equivalent to X=0 for all M, except M=0, which is invalid.

eq2 is identical to eq3, so eq3 is redundant.

Please clarify.

         -s

On Mon, Jul 24, 2017 at 3:15 PM, Stefano Milani <***@enerconsulting.it> wrote:
Dear all,
 
I would like to solve this system of equation:
 
(T)          Ky*x3+T0
(phi)      Kx*x2+u
(eq1)     (-sin(Kx*x2+u-%pi/4)*(-Ky*x3-T0)-bx*x2)/M=0
(eq2)     (cos(Kx*x2+u-%pi/4)*(Ky*x3+T0)-by*x3-M*g)/M=0
(eq3)     (cos(Kx*x2+u-%pi/4)*(Ky*x3+T0)-by*x3-M*g)/M=0
 
 
Where the required term are x2,x3,u
Which command can I use to obtain all solutions?
 
Best regards
 
 
Stefano Milani
 


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