How to resolve this equation $\exp(\frac{-tI}{RC}) = \cos(\omega tI)$ Is this possible to resolve this equation. I m interested in expressing $tI$ in function of $R,$ $C,$ and $w.$ Here is the equation : 
$$\exp \left(\frac{-tI}{RC}\right) = \cos(\omega tI)$$ where $tI$ is between $3T/4$ and $T.$
Thank you very much !  
 A: $$
\exp \left(\frac{-tI}{RC}\right) = \cos\left( \omega tI \right)
$$
$$
2 \exp \left(\frac{-tI}{RC}\right) = \exp\left( j\omega tI \right) + \exp\left( -j\omega tI \right)
$$
$$\text{where } j= \sqrt{-1}.$$
\begin{align}
& \exp\left( \frac{-tI}{RC} \right) = \exp\left( j\omega t \cdot\frac{-tI/(RC)}{j\omega t} \right) \\[8pt]
= {} & \big(\exp(j\omega t)\big)^{-I/(RCj\omega)} \\[8pt]
= {} & \big(\exp(j\omega t)\big)^{jI/(RC\omega)} \quad \text{since $1/j= -j$} \\[8pt]
= {} & a^{jI/(RC\omega)} \quad \text{where this line defines what $a$ is.}
\end{align}
So we have
$$
2a^{jI/(RC\omega)} = a + \frac 1 a.
$$
$$
2a^{1+ jI/(RC\omega)} = a^2 + 1.
$$
To solve this for $a,$ I might begin by trying Newton's method.
A: You wish to express tI in terms of R, C and $\omega$ :
I suggest the follwing
$$exp(\frac{-tI}{RC})= cos(\omega tI)$$
Differenciate both sides with respect to (tI) ; You'll get :
$$-\frac{1}{RC}exp(\frac{-tI}{RC})= -\omega sin(\omega tI)$$
Now replace $exp(\frac{-tI}{RC})$ from previous equation,
$$-\frac{1}{RC} cos(\omega tI)=-\omega sin(\omega tI)$$
$$tan(\omega tI)=\frac{1}{RC\omega}$$
$$tI=\frac{1}{\omega} tan^{-1}(\frac{1}{RC\omega}) $$
Hope this helps....... :-)
