I have the function $f(x)=\cos^{-1}(\frac{1}{2\cos(x)})$ and I have to find its domain. What I know is that the domain of $\cos^{-1}(x)$ is $[-1,1]$ so I think that $\frac{1}{2\cos(x)}$ should be at least $-1$ and $1$ at most. So $-1=0.5/\cos(x)$ gives $x=\frac{2\pi}{3}+2\pi$ and $1=\frac{1}{2\cos(x)}$ gives $x=\pi/3+2\pi$ i.e according to me the domain of the above function $f(x)=\cos^{-1}(\frac{1}{2\cos(x)})$ is $[2\pi/3+2\pi,\pi/3+2\pi]$. Is that right or there is something wrong?
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$\begingroup$Just computing the boundary points where $\frac{1/2}{\cos x}=\pm 1$ doesn't give you much information about what to do with those boundary points.
Instead, it is easier to think: $\frac{1/2}{\cos x}$ is in $[-1,1]$ exactly when $\cos x \le -\frac12$ or $\cos x \ge \frac 12$. At this point stop doing symbolic algebra and instead draw a graph of the cosine function and read off the graph which intervals satisfy $|\cos x|\ge \frac 12$.
$\endgroup$ 0 $\begingroup$Range of secx = $\mathbb{R}$ - (1, -1) $\Longrightarrow$range of $\frac{1}{2}$sec x = (-$\infty$,-$\frac{1}{2}$] $\cup$[$\frac{1}{2}$, $\infty$) domain of cos$^{-1}x$ = $\left[0,\pi\right]$
$\left[0,\pi\right]$$\cap$ {(-$\infty$,-$\frac{1}{2}$] $\cup$[$\frac{1}{2}$, $\infty$)} = $\left[-1,\frac{1}{2}\right]$$\cup$$\left[0,\frac{1}{2}\right]$
secx $\in$ $\left[-1,\frac{1}{2}\right]$$\cup$$\left[0,\frac{1}{2}\right]$$\Longrightarrow$x $\in$$\left[0,\frac{\pi}{2}\right]$$\cup$$\left[\frac{2\pi}{3},\pi\right]$
Range= $\left[0,\frac{\pi}{2}\right]$$\cup$$\left[\frac{2\pi}{3},\pi\right]$
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