Further trig question (1 Viewer)

vernburn · Sep 20, 2020

vernburn said:
Another solution could be: $\tan x$ is $\pi$ periodic. So, $x$ and $3x$ must differ by some multiple of $\pi$ . i.e.
$3x = x +k\pi,\ \ \ k\in\mathbb{Z}$
$2x = k\pi$
$x = \frac{k\pi}{2}$
for the given domain, let $k = 0,1,2,3,4$ :
$x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ . This is where I believe the answers in the textbook got the extra solutions of $\frac{\pi}{2}$ and $\frac{3\pi}{2},$ from.
However, you are correct to exclude them because tan is undefined for those values.
Therefore the answer should be:
$x = 0, \pi, 2\pi$

I just realized that every time there is $x$ change it to $\phi$ . Sorry if I may have caused confusion.

Further trig question (1 Viewer)

vernburn

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