I have this equation:
$$\frac{\arcsin(\sin(\cos(\Delta)\cdot\frac{2\pi\cdot r}{T+\phi}))-\phi}{\frac{2\pi}{T}}=\cos(\Delta)\cdot r$$
And I want to find an effective way to solve for the Delta variable, but I´m not pretty sure if this is possible, because if I move the terms from one side to the other, I can never mix the two Delta variables together, since they always get stuck inside the cos/acos functions.
Example 1:
(for the first member)
$$\Delta=\arccos\left(\frac{\arcsin(\sin(\cos(\Delta)\cdot\frac{2\pi\cdot r}{T}+\phi))}{\frac{2\pi\cdot r}{T+\phi}}\right)$$
Example 2:
(for the second member)
$$\Delta=\arccos\left(\frac{\arcsin(\sin(\cos(\Delta)\cdot\frac{2\pi\cdot r}{T+\phi}))-\phi}{\frac{2\pi/T}{r}}\right)$$
Note that $\arcsin (\sin x) = x$. So the original equation simplifies to the following:
$$ \frac{T}{2\pi} \left [ \cos(\Delta)\cdot \frac{2\pi r}{T + \phi} - \phi \right ] = r \cos(\Delta) $$
You can then isolate the cosine in this expression and take the inverse cosine to get $\Delta$. First multiply by $2\pi/T$ to get the following:
$$ \begin{align} \cos(\Delta)\cdot \frac{2\pi r}{T + \phi} - \phi &= \frac{2\pi r}{T} \cos(\Delta) \end{align}$$
Then subtract the term on the right, add $\phi$ to both sides, and factor out $2\pi r \cos(\Delta)$ to get:
$$ 2\pi r \cos(\Delta) \left[ \frac{1}{T+\phi} - \frac{1}{T} \right] = \phi $$
Finally, dividing both sides by the extra factors we get the desired result:
$$ \cos(\Delta) = \frac{\phi}{2\pi r} \left[ \frac{1}{T+\phi} - \frac{1}{T} \right]^{-1} $$
All that's left is to take the inverse cosine of both sides. You can clean this up a little more by adding those fractions together as well.