Solving $\sin(t+am)=a$ where $-1 \le m \le 1$ What theory or algorithm would I need to research to solve equations such as $\sin(t+am)=a$ (knowing that $-1 \le m \le 1$) for the value of $a$?
My equations may become more complex but have similar properties (and always the constraint on the range of the $m$ variables), such as $\sin(t+\sin(th+am_1)m_2)=a$.
I've played with Wolfram Alpha enough to verify that these functions stay monotonic with $t$ but still can't figure out how to solve them analytically in the real domain.
EDIT: Here is a piece of music using almost exclusively equations like this to directly calculate the time domain audio samples: http://soundcloud.com/full-synthetic/sunshine It's not the greatest but every synth is on that form.
 A: This does not really answer the question, but it's too big for a comment. Using Newton's method to solve $\sin(t+ma) - a = 0$:
Let $f(x)=\sin(t+ma)-a$, so $f'(x)=m\cos(t+ma)$. Then
$$x_{n+1}=x_n - \frac {\sin(t+mx_n)-x_n}{m\cos(t+mx_n)}=x_n - \frac 1 m \tan(t+mx_n)+\frac{x_n}{m}\sec(t+mx_n)$$
Using WolframAlpha to get a series expansion,
\begin{multline}
x_{n+1} = 
-\frac{\tan t}m+\left(\frac{\sec t}m-\tan^2 t\right)x_n
+\tan t \sec^2 t(\cos t - m)x_n^2 \\
+\frac 1 {24} m \sec^4 t (8m \cos (2t)-16m+15\cos t - 3\cos 3t)x_n^3
+\cdots.
\end{multline}
This looks pretty nasty, but if $t$ and $m$ are fixed (not clear from the problem), you can precompute all the coefficients of $x_n^k$, making each step of Newton's method reasonably efficient. If you start out with a decent initial guess (table lookup, if possible), use just a few terms of the expansion for the first step, and add terms as you proceed, it should probably be fast enough. That said, there may well be a more efficient approach to this particular problem.
A: The bisection technique is also viable, and offers a large benefit over newton's iteration: It guarantees a solution.
