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Find the domain of the following function:

$$f(x)=\ln{\left(2\tan{\left(3x+\frac\pi 3\right)}-2\right)}.$$

The first step that I done is that $f(x)$ is more than $0$.

Then I got to: $1<\tan(3x+\fracπ3)$

How to continue?

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Using the trigonometric circle, we have

$$\tan (3x+\pi/3)>1 \iff $$

$$\frac {\pi}{4}+k\pi <3x+\pi/3<\frac {\pi}{2}+k\pi$$

$$\iff \frac {-\pi}{36}+k\pi/3 <x <\frac {\pi}{18}+k\pi/3$$

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  • $\begingroup$ $\tan x$ has period $\pi.$ Only one equation is necessary $\cdots \frac {\pi}{4} + k\pi < 3x + \frac {\pi}{3} < \frac {\pi}{2} + k\pi$ $\endgroup$ – Doug M Apr 2 '18 at 19:52
  • $\begingroup$ @DougM Yes you're right. i fixed it. thanks. $\endgroup$ – hamam_Abdallah Apr 2 '18 at 19:56
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Nitpick: You want the argument/input of $f$ to be positive, not $f$ itself (the logarithm can be negative!).

To tackle this, write the inequality $ 2 \tan(3x - \frac{\pi}{3}) - 2 > 0.$ This yields, after moving a bit of stuff around, $\tan(3x - \frac{\pi}{3}) > 1$. On the interval $(-\pi, \pi)$, $\arctan$ is increasing so we can take the $\arctan$ of both sides without worrying about sign flips. So, we get $3x - \frac{\pi}{3} > \frac{\pi}{4}$ (why?). Now solve for $x$!

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