Given $\cos(t)=\frac12(e^{it}+e^{-it}$), solve for $\cos^{-1}$(x) in terms of logs/roots. The hint says to let $x=\cos(t)$ and $z=e^{it}$. So I started first by substituting:
$x=\frac12(z+1/z)$ --> multiply both sides by 2z
$2xz=z^2+1$
$z^2 -2zx +1 = 0$ --> quadratic formula, solve for z
$z =x ± \sqrt{x^2-1}$
I'm supposed to solve for t now, but I'm not sure how to proceed from here. Any help would be great!
You've done the hard part. Choosing the positive square root, you have $it = \log(x+\sqrt{x^2-1})$ (where $\log = \ln$). It's easy enough to solve for $t$ by multiplying by $-i$.
Note that $x-\sqrt{x^2-1} = \dfrac 1{x+\sqrt{x^2-1}}$, so, indeed, $-it = -\log(x+\sqrt{x^2-1}) = \log(x-\sqrt{x^2-1})$.