Solving a nasty equation involving sines

Question

I'm doing some basic ray optics and I drilled my problem down to an equation like:

$\sin f(x) = r \times \sin g(x)$

where $f$ and $g$ are rational fractions of $x$ and $r$ is a given number (refraction index).

I'm hoping to get a closed form solution for $x$.

My previous problem didn't involve refraction but only reflection and led to a similar equation but without sines $f_o(x) = g_o(x)$ which I was able to solve (the numerator of $f_o(x)-g_o(x)$ being "only" quartic). Hence I am hoping there is a way to somehow get rid of the sines, for example if $f(x)$ and $g(x)$ were equal, I could use the fact that $a \sin u = b \sin u$ can be transformed to $c \sin (u+\phi)=0$, so $u + \phi = 0$, no more sines involved, done. Is there any more trigonometric mambo-jumbo I could use in this spirit when $f(x)$ and $g(x)$ are different?

Answer 1

For me, whenever I came across such equations(especially in heat transfer), I used to solve them by iterative method. The method seems to be lengthy, but it is intuitive and quite easy too.

The most popular method is Newton's iterative method.

$x_{n+1}=x_n−\frac{F(x_n)}{F′(x_n)}$

Here, $F(x) = sin{f(x)}-rsin{g(x)}$. Differentiate with respect to x to get $F'(x)=cosf(x) \times f'(x) - r cosg(x) \times g'(x) $.

Now, you have to start with a probable guess. As your question doesn't define $f(x)$ and $g(x)$, I can't come up with the guess.

Just plug in the guess as $x_n$ in the above equation. You'll get $x_{n+1}$. Plug this value to the equation again. Repeat the process till you get required precision. The more you repeat the step, successive values of $x_{n+1}$ will get more and more closer.