Solving equation in terms of symbols

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The discussion revolves around solving a complex equation involving multiple variables (x, y, z, m) to express m as a function of x, y, and z. The original poster encountered difficulties using MATLAB and Mathematica to derive the four possible solutions for m. They provided a detailed equation and sought guidance on how to approach the problem. Responses highlighted that the equation is essentially a quartic in m, suggesting that expanding and grouping terms could facilitate finding the roots. One participant noted that Mathematica can quickly find the exact solutions, while the original poster experienced issues, possibly due to using an older version of the software. Ultimately, they managed to obtain the results using a different version of MATLAB, indicating that software version compatibility may affect the output.
quin
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Hi friends
I want to solve an equation which is function of x,y,z,m but i want to find answer like this m=... i want my answer(m) to be a function of x, y, z
but when i solve it in MATLAB or mathematica none of them give me the answer
I know that my equation must have 4 answer for "m" in terms of x , y , z

in code of MATLAB my equation is:

-36 + m* (-64 - 24 *m + m^3) - 16* (2 + m)^2 *cos(x) + 8* cos(2*x) - 28* cos(x - y) + 2 *cos(2 *(x - y)) - 64 *cos(y) - 8* (2* m *(4 + m) + (1 + m)* (7 + m)* cos(x)) *cos(y) + 8 *cos(2*y) - 28* cos(x + y) + 2 *cos(2 *(x + y)) - 64 *cos(z) - 8* (2* m* (4 + m) + (14 + m *(8 + m) + 2 *cos(2*x)) *cos(y) + cos(x) *(14 + m *(8 + m) + 8 *(3 + m) *cos(y) + 2 *cos(2*y))) *cos(z) + 8 *(cos(x) - cos(y))^2 *cos(2*z)==0



i want to solve it and find "m" which is in terms of x,y,z

thank for your guidence
 
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But surely you can do it by hand?

$$
-36 + m(-64 - 24 m + m^3) - 16(2 + m)^2 \cos(x) + 8\cos(2x) - 28\cos(x - y)
+ 2\cos(2(x - y)) - 64\cos(y)\\ - 8(2m(4 + m) + (1 + m)(7 + m)\cos(x))\cos(y) + 8\cos(2y) - 28\cos(x + y) + 2\cos(2(x + y)) - 64\cos(z)\\ - 8(2m(4 + m) + (14 + m(8 + m) + 2\cos(2x))\cos(y) + (14 + m(8 + m) + 8(3 + m)\cos(y) + 2\cos(2y))\cos(x))\cos(z)\\ + 8(\cos(x) - \cos(y))^2\cos(2z)=0$$... is basically finding the roots of a quartic in m:
http://en.wikipedia.org/wiki/Quartic_function

... expand the brackets and group the terms.
I know it's painful: use a whiteboard or a large window.
 
Mathematica finds the four exact solutions of this in a few seconds.

m/.Solve[-36 + m*(-64 - 24*m + m^3) - 16*(2 + m)^2*Cos[x] + 8*Cos[2*x] - 28*Cos[x - y] + 2*Cos[2*(x - y)] - 64*Cos[y] - 8*(2*m*(4 + m) + (1 + m)*(7 + m)*Cos[x])*Cos[y] + 8*Cos[2*y] - 28*Cos[x + y] + 2*Cos[2*(x + y)] - 64*Cos[z] - 8*(2*m*(4 + m) + (14 + m*(8 + m) + 2*Cos[2*x])*Cos[y] + Cos[x]*(14 + m*(8 + m) + 8*(3 + m)*Cos[y] + 2*Cos[2*y]))*Cos[z] + 8*(Cos[x] - Cos[y])^2*Cos[2*z] == 0, m]

Each is huge and contains a forest of Cos[], but they are the exact solutions.

Just curious, what did you give Mathematica that did not find the solutions?
 
Bill Simpson said:
Mathematica finds the four exact solutions of this in a few seconds.

m/.Solve[-36 + m*(-64 - 24*m + m^3) - 16*(2 + m)^2*Cos[x] + 8*Cos[2*x] - 28*Cos[x - y] + 2*Cos[2*(x - y)] - 64*Cos[y] - 8*(2*m*(4 + m) + (1 + m)*(7 + m)*Cos[x])*Cos[y] + 8*Cos[2*y] - 28*Cos[x + y] + 2*Cos[2*(x + y)] - 64*Cos[z] - 8*(2*m*(4 + m) + (14 + m*(8 + m) + 2*Cos[2*x])*Cos[y] + Cos[x]*(14 + m*(8 + m) + 8*(3 + m)*Cos[y] + 2*Cos[2*y]))*Cos[z] + 8*(Cos[x] - Cos[y])^2*Cos[2*z] == 0, m]

Each is huge and contains a forest of Cos[], but they are the exact solutions.

Just curious, what did you give Mathematica that did not find the solutions?

thanks for your help
I used "solve[]" too
and also now I test the sentence that you write above but mathematica said:
"a very larg output was generated"
maybe its reason is that the version of my mathematica is older than yours.

however finally I could reach to my result with other version of Matlab and it quickly gave answer

thank you friend
 
quin said:
I test the sentence that you write above but mathematica said:
"a very large output was generated"
maybe its reason is that the version of my mathematica is older than yours.

I have never seen that error before. What version are you using?

$Version

will tell you this.
 
Mathematica 7

thanks for your help
 

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