<s>[INST] Finding the domain and the range of $f(x) = \ln (1-2\cos x).$ Finding the domain and the range of $f(x) = \ln (1-2\cos x).$
My Attempt: 
I arrived at $x < \cos^{-1}(1/2),$ then how can I complete?
 [/INST] the domain of $\ln (z)$ is $z \gt 0$. So we need $1- 2\cos(x) \gt 0$.
so $$-2\cos(x) > -1$$ $$2\cos(x) \lt 1$$ $$\cos(x) < \frac{1}{2}$$
if we graph $\cos(x)$ we see that $\cos(x) \ge \frac{1}{2}$ for $x \in \big[ \frac{-\pi}{3}, \frac{\pi}{3} \big]$ but we need to exclude this to keep $\cos(x) \lt \frac{1}{2}$, so we take these $x$ values out of this domain. But this region occurs periodically so we need to exclude $x \in \big [ 2n\pi - \frac{\pi}{3},2n\pi + \frac{\pi}{3} \big]$ where $n \in \mathbb Z$
so the domain is $$x \in \mathbb R \setminus \big [2n\pi - \frac{\pi}{3},2n\pi + \frac{\pi}{3} \big]$$ 
As for the range: the max value of $f(x)$ occurs when $\cos(x)$ is a minimum, so the minimum value of $\cos(x)$ in the domain is $\cos(x) = -1$, so putting this into $f(x)$, we get $\ln(1-2(-1))$ = $\ln(3)$ As for the lower bound of the range, $cos(x)$ never actually reaches $\frac{1}{2}$ but it does get infintiely closer to $\frac{1}{2}$ so $1-2\cos(x)$ gets closer to 0 and from our knowledge of logarithms and exponentials, exp(z) gets closer to 0 as z gets closer to -infinity, so f(x) will tend to -infinity, therefore the range is $(-\inf,\ln(3)]$ 
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