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I am trying to solve the following:

\[Beta] = ArcCos[((-2 a^2 b^2 c x + 2 b^2 c^3 x - 2 a^2 b^2 r x - 4 a^2 b c r x - 4 a b^2 c r x + 6 b^2 c^2 r x + 4 b c^3 r x - 4 a^2 b r^2 x - 4 a b^2 r^2 x - 2 a^2 c r^2 x - 8 a b c r^2 x + 4 b^2 c r^2 x + 12 b c^2 r^2 x + 2 c^3 r^2 x - 2 a^2 r^3 x - 8 a b r^3 x - 4 a c r^3 x + 8 b c r^3 x + 6 c^2 r^3 x - 4 a r^4 x + 4 c r^4 x - 4 a b^2 c x^2 - 4 a b^2 r x^2 - 8 a b c r x^2 - 4 b^2 c r x^2 - 8 a b r^2 x^2 - 4 b^2 r^2 x^2 - 4 a c r^2 x^2 - 8 b c r^2 x^2 - 4 a r^3 x^2 - 8 b r^3 x^2 - 4 c r^3 x^2 - 4 r^4 x^2 + \[Sqrt](-(c + r)^2 (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x) + 2 (b + r) x Cos[\[Alpha]])^2 Csc[\[Alpha]]^2 ((a - c) (b + r)^2 (a + c + 2 r) (a - c + 2 x) (a + c + 2 (r + x)) - 4 (b + r)^2 (c + r)^2 x^2 Cot[\[Alpha]]^2 - 4 (b + r) (c + r)^2 x (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x)) Cot[\[Alpha]] Csc[\[Alpha]] - (c + r)^2 (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x))^2 Csc[\[Alpha]]^2))) Sin[\[Alpha]]^2)/((c + r)^2 (a^4 + b^4 + 4 b^3 r + 4 b^2 r^2 - 4 b^2 r x - 8 b r^2 x + 4 b^2 x^2 + 8 b r x^2 + 8 r^2 x^2 + 4 a^3 (r + x) - 4 a (r + x) (b^2 + 2 b r - 2 r x) - 2 a^2 (b^2 + 2 b r - 2 (r^2 + 3 r x + x^2)) + 4 (b + r) x (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x)) Cos[\[Alpha]]))]
Solve[\[Alpha] == ArcCos[((-2 a^2 b c^2 x + 2 b^3 c^2 x - 4 a^2 b c r x + 4 b^3 c r x - 2 a^2 c^2 r x - 4 a b c^2 r x + 6 b^2 c^2 r x - 2 a^2 b r^2 x + 2 b^3 r^2 x - 4 a^2 c r^2 x - 8 a b c r^2 x + 12 b^2 c r^2 x - 4 a c^2 r^2 x + 4 b c^2 r^2 x - 2 a^2 r^3 x - 4 a b r^3 x + 6 b^2 r^3 x - 8 a c r^3 x + 8 b c r^3 x - 4 a r^4 x + 4 b r^4 x - 4 a b c^2 x^2 - 8 a b c r x^2 - 4 a c^2 r x^2 - 4 b c^2 r x^2 - 4 a b r^2 x^2 - 8 a c r^2 x^2 - 8 b c r^2 x^2 - 4 c^2 r^2 x^2 - 4 a r^3 x^2 - 4 b r^3 x^2 - 8 c r^3 x^2 - 4 r^4 x^2 + \[Sqrt]((b + r)^2 (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x) + 2 (c + r) x Cos[\[Beta]])^2 Csc[\[Beta]]^2 (-((a -  b) (c + r)^2 (a + b + 2 r) (a - b + 2 x) (a + b +  2 (r + x))) + 4 (b + r)^2 (c + r)^2 x^2 Cot[\[Beta]]^2 + 4 (b + r)^2 (c + r) x (a^2 - c^2 - 2 c r + 2 r x +   2 a (r + x)) Cot[\[Beta]] Csc[\[Beta]] + (b + r)^2 (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x))^2 Csc[\[Beta]]^2))) Sin[\[Beta]]^2)/((b +  r)^2 (a^4 + c^4 + 4 c^3 r + 4 c^2 r^2 - 4 c^2 r x - 8 c r^2 x + 4 c^2 x^2 + 8 c r x^2 + 8 r^2 x^2 + 4 a^3 (r + x) - 4 a (r + x) (c^2 + 2 c r - 2 r x) - 2 a^2 (c^2 + 2 c r - 2 (r^2 + 3 r x + x^2)) + 4 (c + r) x (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x)) Cos[\[Beta]]))], \[Alpha]]

I realise that these are complex expressions, and it makes sense that Wolfram Mathematica / my computer is having a hard time, but I need to know if I am going to wait a few more hours or a few more years.

QUESTIONS:

Is there anyway I could get a ETA for when the evaluation will be done?

Do you have any suggestions to speed up the process?

Is there perhaps a strong VPS/ remote desktop service you can recommend that will be able to solve this faster?

I see Mathematica is only using 25% of my CPU, can i increase this?

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    $\begingroup$ I think it's unlikely that you will get a useful answer from this. If it takes a very long time, it is likely to either fail or return an expression that has an enormous number of terms. Such expressions are rarely useful. What would you do with it? More practical ideas: is a numerical solution acceptable? might some terms be very small, will an approximation be good enough? $\endgroup$
    – mikado
    Jun 19 at 15:10
  • $\begingroup$ At a minimum you should try to simplify the expression using any known constraints/assumptions prior to trying to Solve. Are any parameters/variables real? Nonnegative? Positive? Constrained to intervals (particularly periodic variables)? Also, include constraints/assumptions in Solve. $\endgroup$
    – Bob Hanlon
    Jun 19 at 15:35
  • $\begingroup$ @mikado Thanks for your comment! Unfortunately an estimation would not suffice. I am pretty sure an answer will be useful. The appalication is a high accuracy measuring machine. The machine has 3 measuring probes at 0° α° and β°. I have 3 equations and 3 unknowns. x(the final measurment), α and β. I am trying to do simultaneous equation. therefore I need an expression for α in terms of β. $\endgroup$
    – Msegling
    Jun 19 at 15:39
  • $\begingroup$ @Msegling I think you are most likely to make progress by giving numerical values to as many of your constants as possible. $\endgroup$
    – mikado
    Jun 19 at 15:48
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I couldn't get a result reasonably fast, but in general you can make Solve usually much faster by replacing the trigonometric functions by variables and use ArcTan[cos, sin] subsequently. That is:

eq1 = cosβ == ((-2 a^2 b^2 c x + 2 b^2 c^3 x - 2 a^2 b^2 r x - 4 a^2 b c r x - 4 a b^2 c r x + 6 b^2 c^2 r x + 4 b c^3 r x - 4 a^2 b r^2 x - 4 a b^2 r^2 x - 2 a^2 c r^2 x - 8 a b c r^2 x + 4 b^2 c r^2 x + 12 b c^2 r^2 x + 2 c^3 r^2 x - 2 a^2 r^3 x - 8 a b r^3 x - 4 a c r^3 x + 8 b c r^3 x + 6 c^2 r^3 x - 4 a r^4 x + 4 c r^4 x - 4 a b^2 c x^2 - 4 a b^2 r x^2 - 8 a b c r x^2 - 4 b^2 c r x^2 - 8 a b r^2 x^2 - 4 b^2 r^2 x^2 - 4 a c r^2 x^2 - 8 b c r^2 x^2 - 4 a r^3 x^2 - 8 b r^3 x^2 - 4 c r^3 x^2 - 
        4 r^4 x^2 + √(-(c + r)^2 (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x) + 2 (b + r) x Cos[α])^2 Csc[α]^2 ((a - c) (b + r)^2 (a + c + 2 r) (a - c + 2 x) (a + c + 2 (r + x)) - 4 (b + r)^2 (c + r)^2 x^2 Cot[α]^2 - 4 (b + r) (c + r)^2 x (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x)) Cot[α] Csc[α] - (c + r)^2 (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x))^2 Csc[α]^2))) Sin[α]^2)/((c + r)^2 (a^4 + b^4 + 4 b^3 r + 4 b^2 r^2 - 4 b^2 r x - 8 b r^2 x + 4 b^2 x^2 + 8 b r x^2 + 8 r^2 x^2 + 4 a^3 (r + x) - 4 a (r + x) (b^2 + 2 b r - 2 r x) - 2 a^2 (b^2 + 2 b r - 2 (r^2 + 3 r x + x^2)) + 4 (b + r) x (a^2 - b^2 - 2 b r + 2 r x + 2 a (r + x)) Cos[α]));
eq2 = cosα == ((-2 a^2 b c^2 x + 2 b^3 c^2 x - 4 a^2 b c r x + 4 b^3 c r x - 2 a^2 c^2 r x - 4 a b c^2 r x + 6 b^2 c^2 r x - 2 a^2 b r^2 x + 2 b^3 r^2 x - 4 a^2 c r^2 x - 8 a b c r^2 x + 12 b^2 c r^2 x - 4 a c^2 r^2 x + 4 b c^2 r^2 x - 2 a^2 r^3 x - 4 a b r^3 x + 6 b^2 r^3 x - 8 a c r^3 x + 8 b c r^3 x - 4 a r^4 x + 4 b r^4 x - 4 a b c^2 x^2 - 8 a b c r x^2 - 4 a c^2 r x^2 - 4 b c^2 r x^2 - 4 a b r^2 x^2 - 8 a c r^2 x^2 - 8 b c r^2 x^2 - 4 c^2 r^2 x^2 - 4 a r^3 x^2 - 4 b r^3 x^2 - 8 c r^3 x^2 - 
        4 r^4 x^2 + √((b + r)^2 (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x) + 2 (c + r) x Cos[β])^2 Csc[β]^2 (-((a - b) (c + r)^2 (a + b + 2 r) (a - b + 2 x) (a + b + 2 (r + x))) + 4 (b + r)^2 (c + r)^2 x^2 Cot[β]^2 + 4 (b + r)^2 (c + r) x (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x)) Cot[β] Csc[β] + (b + r)^2 (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x))^2 Csc[β]^2))) Sin[β]^2)/((b + r)^2 (a^4 + c^4 + 4 c^3 r + 4 c^2 r^2 - 4 c^2 r x - 8 c r^2 x + 4 c^2 x^2 + 8 c r x^2 + 8 r^2 x^2 + 4 a^3 (r + x) - 4 a (r + x) (c^2 + 2 c r - 2 r x) - 2 a^2 (c^2 + 2 c r - 2 (r^2 + 3 r x + x^2)) + 4 (c + r) x (a^2 - c^2 - 2 c r + 2 r x + 2 a (r + x)) Cos[β]));

rules = Thread[{Cos[α], Cos[β], Cot[α], Cot[β], Csc[α], Csc[β], Sin[α], Sin[β]} ->
                 {cosα, cosβ, cosα/sinα, cosβ/sinβ, 1/sinα, 1/sinβ, sinα, sinβ}];

Solve[{eq1 /. rules, eq2 /. rules,
       cosα^2 + sinα^2 == 1, cosβ^2 + sinβ^2 == 1}, {cosα, sinα, cosβ, sinβ}]
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  • $\begingroup$ Thank you for your answer. I will try this. Do you know of any remote kernel services? Or is there alternatively a service I can pay for to have my equations solved? I have tried running Mathematica on a powerful VPS but It doesn’t seem to utilise the full potential of the machine. When you run mathematica on your system do you use near 100% of your CPU and Memory? $\endgroup$
    – Msegling
    Jun 20 at 10:50

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