Question 18605
Hello There:

I'll take a stab at it.

P2/P1 = cos[(L*pi)/(2*X)]^2

Take the square root of both sides.

cos[(L*pi)/(2*X)] = +/- sqrt(P2/P1)

Use the inverse cosine function (arccos) to get L out of the input to cosine.

arccos[+/- sqrt(P2/P1)] = L*pi/(2*X)

Multiply both sides by (2*X)/pi.

L = [(2*X)/pi]*arccos[+/- sqrt(P2/P1)]

Hopefully, this matches the result in your book.  You need to decide what to do about the +/- part.  In other words, maybe only the positive one is applicable.

~ Mark