I Cannot Solve This Algebra-Trig Equation If someone can guide me on how to solve this, that'll be great:
$$\frac{4805}{a} = \cos^{-1} \left(\frac{a-4.2}{a}\right)$$
solve for $a$.
Cheers
 A: This is not an equation that can be solved algebraically. The best way to solve it is to plug it into a computer and ask for the answer. If you need to calculate it by hand, there are some approximations of $\arccos$ that converge reasonably quickly that can be used, and if you want I can provide them.
A: This is a  transcendental equation and the solution can be obtained only by numerical method (notice that this already the case for the simple $x=\cos(x)$).
Taking into account the fact that $-1\leq\cos(x)\leq 1$, it is clear that the zero of equation $$f(a)=\frac{4805}{a}-\cos ^{-1}\left(\frac{a-\frac{21}{5}}{a}\right)$$ must be very large.
To get a first approximation, let us use Taylor series. This will give $$f(a)=-\sqrt{\frac{42}{5}} {\frac{1}{a^{1/2}}}+\frac{4805}{a}+O\left(\frac{1}{a^{3/2}}\right)$$ which gives as an estimate $$a_0=\frac{115440125}{42}\approx 2.74857\times 10^6$$ Now, we can start Newton method which will provide the following successive iterates
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 2.7485744047619047619\times 10^6 \\
 1 & 2.7485737047613966799\times 10^6 \\
 2 & 2.7485737047616195229\times 10^6 
 \end{array}
\right)$$ which is the solution for twenty significant figures.
Edit
May be simpler, we could consider the problem to be : find the zero of equation
$$g(x)=\frac{21 }{5}x+\cos (4805 x)-1$$ where $x=\frac 1a$. Its derivative is $$g'(x)=\frac{21}{5}-4805 \sin (4805 x)$$ which cancels for $$x_*=\frac{1}{4805}\sin ^{-1}\left(\frac{21}{24025}\right)$$ This corresponds to a maximum (by the second derivative test). Since $g(0)=0$, the solution is such that $x > x_*$.
Now, using Taylor expansion again around $x=0$ 
$$g(x)=\frac{21 }{5}x-\frac{23088025}{2} x^2+O\left(x^3\right)$$ which cancels for $$x=\frac{42}{115440125}\implies a=\frac{115440125}{42}$$ 
