If

\(\begin{cases} \tan { x } \tan { y } = \frac { \sin { z } }{\cos{x}\cos{y} } +3 \\ \tan{y}\tan{z}=\frac {\sin{x}}{\cos{y}\cos{z}}-5 \\ \tan{z}\tan{x}=\frac {\sin{y}}{\cos{z}\cos{x}}-3 \end{cases},\)

and

\(x=a+bm;\) \(g+hm\),

\(y=c+dn;\) \(i+jn\),

\(z=e+f(2p-m-n);\) \(k+l(2p-m-n+1)\).

where \(a,b,c,d,e,f,g,h,i,j,k,\) and \(l\) are constants, and \(n,m,p\in \mathbb{Z} \),

evaluate

\(\left| \left\lfloor a\cdot b\cdot c\cdot d\cdot e\cdot f\cdot g\cdot h\cdot i\cdot j\cdot k\cdot l+\pi \right\rfloor \right| \).

DETAILS AND ASSUMPTIONS:

\(m,\) \(n,\) and \(p,\) represent the multiplicities of periods of solutions. Ex: If \(\sin{(x)}=0\), then \(x=\pi n, n\in \mathbb{Z}\).

\(m,\) \(n,\) and \(p,\) encompass all integers. So \(2p\) would go: \(...,-2,0,2,4,...\).

Notice that in the presented form of solutions, constants like \(b,\) \(d,\) and \(f\) are uniquely determined. While \(f*(2p)=(2f)*p,\) as \((2f)*p\) \(f\) would not factor out nicely with \(m\) and \(n\)... etc.

If any constant turns out to be 0 (one of them does), please treat it as 1. I've just noticed this, so if anyone entered 3, you're right. Sorry about that!

Please enter the solution expansion points based on the domain of the principal value of the relevant trigonometric function (defined by the range of its inverse function). For example, for \(\sin{(x)}=-1\), \(x=-\frac{\pi}{2}+2\pi n\) (the principal range of \(\sin^{-1}{(x)}\) is \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)).

[HINT: all constants \(\leq \pi\).]

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