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What would be the correct approach solve this complicated equation in Mathematica?

Posted 19 days ago

I am just getting started with Mathematica and I have become familiar with quite a few basic operations. I wanted some advice on solving a complicated equation which I have encountered.

This is the equation I am trying to solve for $\alpha1^{\circ}$: $$ \text{c1} \left(\frac{1}{\cos (\text{$\alpha $1} {}^{\circ}) \sqrt{1-\text{M1}^2 \sin ^2(\text{$\alpha $1} {}^{\circ})}-\text{M1} \sin ^2(\text{$\alpha $1} {}^{\circ})}+\text{M1}\right) = \text{$\alpha $2} (\text{M2}+1)$$.

In this equation, $M1$, $M2$, $c1$ and $c2$ are all non-zero and non-negative constants. I have worked out by hand the solution of this equation but I want to understand how I can do this in Mathematica. This is the approach which I have used:

  1. In the complicated equation let $$ Y = \cos (\text{$\alpha $1} {}^{\circ}) \sqrt{1-\text{M1}^2 \sin ^2(\text{$\alpha $1} {}^{\circ})} - \text{M1} \sin ^2(\text{$\alpha $1} {}^{\circ}) $$ This then simplifies the equation of interest to $$ \text{c1} \left(\text{M1}+\frac{1}{\text{Y}}\right) = \text{$\alpha $2} (\text{M2}+1)$$

  2. The last simplified equation can be solved for $Y$ which yields $$ Y = \frac{\text{c1}}{-\text{c1} \text{M1}+\text{c2} \text{M2}+\text{c2}} $$ Since the solution $Y$ is composed entirely of constants, the final solution will also be a constant. Thus, let $$ K = \frac{\text{c1}}{-\text{c1} \text{M1}+\text{c2} \text{M2}+\text{c2}} $$. With this simplification the equation of interest becomes $$ Y = K$$ Substituting the simplified form of $Y$ from Step 1) leads to $$ \cos (\text{$\alpha $1} {}^{\circ}) \sqrt{1-\text{M1}^2 \sin ^2(\text{$\alpha $1} {}^{\circ})} - \text{M1} \sin ^2(\text{$\alpha $1} {}^{\circ}) = K $$

  3. In $Y$, let $$ x = \sin^{2}(\alpha1)$$ then $Y$ in terms of $x$ becomes $$ Y = \sqrt{1-x} \ \ \sqrt{1-\text{M1}^2 x}-\text{M1} x $$

  4. The last equation can be solved for $x$ and after substituting the relevant expressions the final answer will be given in terms of $\sin^{-1}$ and there will be four solutions of $\alpha1$.

These are the four major steps I have used to manually solve the equation. Can someone please advise on what would be right approach to solve this in Mathematica ? What would be the right approach? I have attached here the notebook with the LHS and RHS parts of the equation typed in.

POSTED BY: Ishan Nande
3 Replies

I get four solutions with Solve this way:

sols = Solve[LHS == RHS, \[Alpha]1]

The solutions are to be verified. Here is how they behave with a choice of the parameters:

paramValues = {c1 -> 1, c2 -> 1, M1 -> 2, M2 -> 3};
sols /. paramValues
% // N
LHS == RHS /. sols /. paramValues
POSTED BY: Gianluca Gorni
Posted 19 days ago

Yes. I corrected it.

POSTED BY: Ishan Nande

I think (1-x^2) should actually be sqrt(1-x) near the end.

POSTED BY: Daniel Lichtblau
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