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A problem in Optics comes down to this equation:

Given that $A$, $B$, $C$ and $D$ are all real positive numbers, $-1 \lt x \lt 1$ and $-1 \lt \sqrt{C} x \lt 1$

$$\frac{Ax}{\sqrt{1 - x^2}} + \frac{Bx}{\sqrt{1 - Cx^2}} = D$$

How to solve this kind of equation?

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    $\begingroup$ To use a dot for multiplication, use \cdot, so A\cdot x gives $A \cdot x$, or you can use nothing. $\endgroup$ – Ross Millikan Aug 11 '16 at 21:10
  • $\begingroup$ Is it related to THIS? $\endgroup$ – Ng Chung Tak Aug 12 '16 at 15:20
  • $\begingroup$ @NgChungTak yes definitely, that is exactly what I am looking for $\endgroup$ – ielyamani Aug 12 '16 at 15:32
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You multiply by the product of the denominators and square the result. That will get you down to one square root, which you can isolate and square again. Unfortunately, that gives you an eighth degree equation unless there is cancellation. You can do numeric root finding on the original.

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  • $\begingroup$ The eighth degree equation will be a quartic equation in $x^2$, so theoretically you could end up with eight exact though messy solutions, some of which may be spurious $\endgroup$ – Henry Aug 11 '16 at 21:13
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Firstly I think that either the second denominator is $\sqrt{1-(Cx)^2}$ or the condition with $C$ is $-1 \leq \sqrt{C}x \leq 1$ with the additional $C \geq 0$.

A possible approach to the equation is the substitution $x = \cos{y}$ ot $x = \sin{y}$ (since $|x|<1$). With the first substitution, the equation becomes $ \frac{B\cos{y}}{\sqrt{1-C\sin^2{y}}} = D - A\cot{y}$. Now a necessary condition for the equation to have a solution is $(D-A\cot{y})(B\cos{y}) \geq 0$. For all such $y$ rise to the power of two and you will get rid of the square root. However, the left trigonometric equation could be difficult to solve explicitly for arbitrary parameters $A,B,C,D$. You can use universal substitution to make the equation polynomial. If you obtain some roots remember check if for them holds $(D-A\cot{y})(B\cos{y}) \geq 0$. It is possible that the problem must be solved numerically after the last step with the universal substitution.

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