Finding the domain and the range of $f(x) = \ln (1-2\cos x).$

Question

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?

Answer 1

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 $(-\inf,\ln(3)]$

Answer 2

$$1-2\cos x>0\implies \cos x <{1\over 2}$$

The solutions for $\cos x ={1\over 2}$ is $$x=\pm {\pi\over 3} +2\pi\cdot k$$ where $k\in \mathbb{Z}$. Thus, if you draw a graph for $y=\cos x$ you see that it is under the line $y={1\over 2}$ for $$x\in \bigcup _{k\in \mathbb{Z}}( {\pi\over 3} +2\pi k, {5\pi\over 3} +2\pi k)$$ and thus this is a domain of $f$.

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Now do the range: Since we have $$\lim _{x\to {\pi\over 3}} \ln(1-2\cos x) = -\infty$$ and $f(x)= \ln 3$ for $x=\pi$ we have $$R_f = (-\infty, \ln 3]$$

since $f$ is continous on $[{\pi\over 3}, {5\pi\over 3}]$

Answer 3

You are using the $ln$ function therefore $1-2cosx$ must be strictly positive, hence solve the inequation $$1-2cosx>0$$

or for which condition on $x$, we have $cos(x)<\frac{1}{2}$. Here

$$x\in \bigcup _{n\in \mathbb{Z}}] {\frac{\pi}{3}} +2\pi n, \frac{5\pi}{3} +2\pi n[$$

Answer 4

Let $y = \ln(1-2\cos x)\implies e^y = 1- 2\cos x\implies |\cos x |= \left|\dfrac{1-e^y}{2}\right|\le 1\implies -2 \le 1-e^y \le 2\implies e^y \le 3\implies y \le \ln 3\implies \text{range} = (-\infty,\ln 3]$. Domain is the same as in the other posts.