I was trying to find the roots to a trigonometric equation and so I asked a question in this mathematics server/forum. Later, someone posted an answer: basically, approached it using perturbation theory. He got an approximation of its roots until $\epsilon^1$, namely, in the form of $a_{o}+\epsilon\ a_{1}$. I could only get to $a_{o}$, thus I just wanted someone to help me get to at least $a_{1}$ or $a_{2}$. I'll write the equation I'm trying to solve and the $a_{n}$ terms.
Trig. equation: $$\sin{\xi \Psi}-\zeta \frac{\cos{\xi \Psi}}{\cos{\Psi}}=0,$$ where $0≤\xi≤1$, $0≤\zeta≤2$ or $3$, and $0<\Psi≤ \frac{\pi}{2}.$
Perturbed equation:
If you let $\xi=1-\epsilon\ $ and $\ \Psi(\epsilon)=\sum\limits_n \epsilon^n a_n$, the previous equation reduces to this:
$$\sin{(1-\epsilon) \Psi}-\zeta \frac{\cos{(1-\epsilon) \Psi}}{\cos{\Psi}}=0.$$
$a_{0}:$ $$a_{0}= \frac{\pi}{2}\ \mathrm{or}\ \arcsin{\zeta}$$
$a_{1}:$ $$a_1=a_o \frac{\cos^2(a_o)+\zeta\sin(a_o)}{\cos^2(a_o)+(\zeta-1)\sin(a_o)}$$