I'm lost with the process to solve this equation.
$$38\sin\left(x-57.5\right)-66.6e^{-(\pi x)/(180\,\cdot\,1.57)}=30.$$
The exponential value for $x$ is in radians, thus I've converted to degrees. I have an answer of $200.5$ degrees, I'm just not sure how to get there.
The original question was
(c) the extinction angle $\beta$ of the current (i.e. when it reaches zero after the input ac voltage becomes negative). Compare this angle to the pf angle of the RL load. ($200.55^\circ$)
ANS: Solve $i(\omega t) = 38\sin\left(\omega t - 57.5^\circ\right) + 66.57 e^{-\omega t/1.57} - 30 = 0$
numerically $\to \omega t = \beta = 200.55^\circ$
(screen shot: https://i.stack.imgur.com/CWjg3.jpg )
Here's the original equation with the substitution $x = \omega t$ and with all angle values consistently stated in radians, which is the natural way to deal with angles in problems like this: $$ 38\sin(x - 1.0036) + 66.57 e^{-x/1.57} - 30 = 0. $$
An equation like this has to be solved numerically. Basically this comes down to "guess and check." Newton-Raphson is just a sophisticated guess-and-check method with some formulas to help you make very good guesses.
The answer key claims the answer is $x = 3.5003$ (that is, $200.55^\circ$ converted to radians). Let's try this: \begin{align} 38\sin(3.5003 - 1.0036) &\approx 22.84 \\ 66.57 e^{-3.5003/1.57} &\approx \phantom0 7.16 \\ 38\sin(3.5003 - 1.0036) + 66.57 e^{-3.5003/1.57} &\approx 30 \end{align}
That looks about right.
You will not get this solution from the equation you wrote, however, because you changed one of your $+$ symbols to $-$ while you were converting things to degrees.
You can see other solutions of the equation by entering it in Wolfram Alpha:
https://www.wolframalpha.com/input/?i=38sin(x-1.003564)%2B66.57e%5E(-x%2F1.57)-30%3D0
Interestingly, this gives two more roots near $x = 0.438.$ This may be an artifact of rounding that was performed when writing the equation. Even if it is not, the value of $i(\omega t)$ barely goes negative and does not stay negative very long at all. So there is a question (which can only be answered by digging deeper into the original context of the formula) about why the lesser of these roots is not considered to be an "extinction angle".