Solving Fiber Optics Glass Characteristic Equations with Maple

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The discussion centers on solving the characteristic equations of fiber optics glass using Maple software. The user is struggling to obtain solutions for the propagation angle x, as the equations yield similar expressions for different conditions (Pi and 0). They mention that while Maple is unhelpful, their TI calculator provided similar results that seem incorrect. The equations represent different modes of propagation, and the user seeks clarity on how to present the problem more effectively to attract assistance. The general equation involves parameters like refractive indices and wavelength, suggesting a need for trigonometric identities to simplify the problem.
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Homework Statement


They ask me to solve the characteristic equation of a fiber optics glass. The equations I need to solve (separately) are:


Homework Equations



solve(2000000*Pi*sin(x)-2*arctan(.6666666667*(2.25*cos(x)^2-2.1904)^.5/sin(x)) = Pi, x):

solve(2000000*Pi*sin(x)-2*arctan(.6666666667*(2.25*cos(x)^2-2.1904)^.5/sin(x)) = 0, x)

That's how I tried to solve them in MAPLE.



The Attempt at a Solution



Maple is not giving me answers, I used my TI and it gave me some results but they were extremely similar and I don't think they are the real ones. How can I solve for the propagation angle x?
 
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The expressions on the left look the same. How can the same expression be both Pi and 0 at the same time?

And please try using tex or at least 2*10^6 and 2/3 instead of 2000000 and .6666666667
More people may decide to help if you put effort in presenting your problem neatly.
 
Well I'm Maple I used fractions, when I copy pasted them here they appeared as decimals. The equations are supposed to be different modes or propagation, so I need to solve for each angle individually.

The equation (general) is:

m*PI=2*PI*d/lambda -2 arctan(SQRT((n12Cos(x))^2 -n2^2)/n1*sin(x))

Where N1=1.5
n2=1.48
lambda=1(10^-6)
d=(3.192(10^-6)
And m=0,1,2,3... Where I only need the first (0 and 1)
 
If you define
\alpha^2 =1-(n_2/n_1)^2
and let
y=sin(x)/\alpha, then I think you can write the arctan as:
arctan\left(\frac{\sqrt{1-y^2}}{y}\right)

Try some trig identities from there.
 
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