is arctan(-pi/3) same thing as arctan(2pi/3)

and arctan(-4pi/3) same thing as arctan(5pi/3)?

if not, why is it that one?

hence solve $\displaystyle tanx=-\sqrt{3}$

thanks

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- Mar 1st 2010, 11:42 AM #1

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- Mar 1st 2010, 12:10 PM #2
In the case of

$\displaystyle \tan(x) =-\sqrt{3} $

In the first snd and 4th quadrants $\displaystyle x = \pi - \frac{\pi}{3} , 2\pi - \frac{\pi}{3}= \frac{2\pi}{3} , \frac{5\pi}{3}$

As $\displaystyle \tan(x)$ has the period $\displaystyle \pi $ in general you can say

$\displaystyle x = \frac{2\pi}{3} \pm n\pi , n = 0,1,2,\dots $

- Mar 1st 2010, 12:46 PM #3

- Mar 1st 2010, 01:01 PM #4

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- Mar 1st 2010, 01:31 PM #5

- Mar 1st 2010, 06:14 PM #6

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$\displaystyle tan\left(-\frac{\pi}{3}\right)=tan\left(\frac{2{\pi}}{3}\rig ht)$

$\displaystyle tan\left(-\frac{4{\pi}}{3}\right)=tan\left(\frac{5{\pi}}{3}\ right)$

$\displaystyle tan\left(\frac{\pi}{3}\right)=\sqrt{3}$

$\displaystyle tan(-A)=-tanA$

$\displaystyle tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}$

$\displaystyle tan\left(-\frac{\pi}{3}\right)=tan\left(\frac{2{\pi}}{3}\rig ht)$

Therefore, if

$\displaystyle tanx=-\sqrt{3}$

then

$\displaystyle x=arctan\left(\frac{2{\pi}}{3}\right)$

- Mar 5th 2010, 10:37 AM #7

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