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terryhendicottAnswer
21-Topaz II
June 4, 2023

Hi Jason,

If every variable except Tr2 is defined you could use the root() function or a solve block.

Cheers

Terry

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terryhendicott21-Topaz IIAnswer
21-Topaz II
June 4, 2023

Hi Jason,

If every variable except Tr2 is defined you could use the root() function or a solve block.

Cheers

Terry

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J_power11-GarnetAuthor
11-Garnet
June 5, 2023

Hi @terryhendicott ,

 

Yes, every variable except Tr2 is defined.

I try to follow your method and I get the result.

Jason_power_0-1685965693534.png

but, why Mathcad can't help to arrange the equation if there is a sine in there.

 

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Werner_E
Werner_E25-Diamond I
25-Diamond I
June 5, 2023
I try to follow your method and I get the result.

 

If a numerical result is desired, its probably better to use numerical methods like the "root" function or a solve block with "Find".

 

but, why Mathcad can't help to arrange the equation if there is a sine in there.

 

Again, why do you think that Mathcad should be able to do?

Apart from the fact that the symbolics in Mathcad and even worse in Prime is not state of the art compared to other current software products, its also a fact that by far not all mathematical equation have a closed, symbolic solution.  Actually most haven't 😉

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Werner_E
Werner_E25-Diamond I
25-Diamond I
June 4, 2023

Why do you think that a generic symbolic solution would even exist?

You won't even find a symbolic solution for a much simpler equation like sin(x+1)=a*x

Werner_E_0-1685885996170.png

 

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J_power11-GarnetAuthor
11-Garnet
June 5, 2023

Hi @Werner_E 

 

Because I think the answer exists, and I can get a math equation.

 

Now, I understand! thanks for your explanation.

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Werner_E
Werner_E25-Diamond I
25-Diamond I
June 5, 2023

O don't think that a closed symbolic solution for your equation exists, but if it does, it sure would have to contain a myriad of "if's" concerning the other variables and their values with respect to each other.

 

The simple equation sin(x)=a*x+b can have (depending on a) none to any finite number of solutions (but not infinite!) if b>1 or b<-1, but it can have at least 1 up to infinite solutions if -1<=b<=1.

How would all these cases be expressed in one single symbolic solution?

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