Question:

How to analytically solve this equation with trigonometric function

Abigail: 26 June 2022

The equation containing trigonometric function is:

R == Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]*Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]*c/Sqrt[(Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]+Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]+c)*(-Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]+Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]+c)*(Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]-Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]+c)*(Sqrt[(i1-Sin[t])^2 + j1^2 + (k1-Cos[t])^2]+Sqrt[(i2-Sin[t])^2 + j2^2 + (k2-Cos[t])^2]-c)]

The t falls in range of (0,Pi/2). I want to solve t. How to make it?

This equation is too complicated and my computer could solve it for a long time?

Answer:
Easton: 26 June 2022

This is your equation:

eq1 = R == Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2]*
    Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2]*
    c/Sqrt[(Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] + 
         Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] + 
         c)*(-Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] + 
         Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] + 
         c)*(Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] - 
         Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] + 
         c)*(Sqrt[(i1 - Sin[t])^2 + j1^2 + (k1 - Cos[t])^2] + 
         Sqrt[(i2 - Sin[t])^2 + j2^2 + (k2 - Cos[t])^2] - c)];

Let us transform it as follows:

eq2 = eq1 /. Sin[t] -> Sqrt[1 - Cos[t]^2] /. Cos[t] -> x

That gives a rather long expression. For this reason I do not write it here. It is in terms of x=Cos[t]. If we get x, we can then calculate arccos(x), provided -1<x<1. OK, let us solve it:

Solve[eq2, x]

It is solved exactly in somewhat about 10 min, but the result is a huge analytic expression, such that one can hardly make use of it.

Depending on the nature of the problem in such a situation I would look for some sort of a simplification of this equation.

Have fun!