# How to solve this equation with square roots and fractions?

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?

• To use a dot for multiplication, use \cdot, so A\cdot x gives $A \cdot x$, or you can use nothing. – Ross Millikan Aug 11 '16 at 21:10
• Is it related to THIS? – Ng Chung Tak Aug 12 '16 at 15:20
• @NgChungTak yes definitely, that is exactly what I am looking for – ielyamani Aug 12 '16 at 15:32

• 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 – Henry Aug 11 '16 at 21:13
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.