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March 29th, 2016, 02:09 PM  #1 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3  Why can't I add the period to find the next two solutions?
Consider tan(2x) = $\displaystyle \sqrt3$ for 0 $\displaystyle \leq x < 2\pi$ reference angle x = $\displaystyle \tan^{1}$($\displaystyle \sqrt{3}$) = $\displaystyle \frac{\pi}{3}$ In the given domain 0 $\displaystyle \leq x < 2\pi$, the positive ratio of tangent is in quadrants I and III. So the two standard position angles in this interval are 2x = $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{4\pi}{3}$ Now, divide both sides by 2. Then I get x = $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{2\pi}{3}$  (*) The period of tangent is: period = $\displaystyle \frac{\pi}{b}$ where b = 2 in tan(2x). So the period is $\displaystyle \frac{\pi}{2}$. Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle. So I get x = $\displaystyle \frac{2\pi}{3}$ and $\displaystyle \frac{7\pi}{6}$ Therefore, the solutions are x=$\displaystyle \frac{\pi}{6}$, $\displaystyle \frac{2\pi}{3}$, $\displaystyle \frac{2\pi}{3}$ and $\displaystyle \frac{7\pi}{6}$ The two middle solutions are the same. The answer key answer says x = $\displaystyle \frac{\pi}{6}$, $\displaystyle \frac{2\pi}{3}$, $\displaystyle \frac{7\pi}{6}$ and $\displaystyle \frac{5\pi}{3}$. I actually found my mistake. If I added $\displaystyle \pi$ instead of $\displaystyle \frac{\pi}{2}$ in (*) above, I would get the right solutions in the second cycle. But, I don't understand why you have to do that, since the period of tan(2X) is $\displaystyle \frac{\pi}{b}$ = $\displaystyle \frac{\pi}{2}$. I am very confused. Can someone explain this? Thanks a lot. Last edited by skipjack; March 29th, 2016 at 06:06 PM. 
March 29th, 2016, 02:40 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 
$\tan(2x) = \sqrt{3}$ $0 \le x < 2\pi \implies 0 \le 2x < 4\pi$ $2x = \dfrac{\pi}{3} \, , \, \dfrac{4\pi}{3} \, , \, \dfrac{7\pi}{3} \, , \, \dfrac{10\pi}{3}$ $x = \dfrac{\pi}{6} \, , \, \dfrac{2\pi}{3} \, , \, \dfrac{7\pi}{6} \, , \, \dfrac{5\pi}{3}$ note ... $\dfrac{\pi}{6} + \dfrac{\pi}{2} = \dfrac{2\pi}{3}$ $\dfrac{\pi}{6} + 2 \cdot \dfrac{\pi}{2} = \dfrac{7\pi}{6}$ $\dfrac{\pi}{6} + 3 \cdot \dfrac{\pi}{2} = \dfrac{5\pi}{3}$ ... adding the modified period does work 
March 29th, 2016, 02:42 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 
Note that the difference between $\dfrac{\pi}{6}$ and $\dfrac{2\pi}{3}$ is $\dfrac{\pi}{2}$. If you had added $\dfrac{\pi}{2}$ to $\dfrac{2\pi}{3}$ twice you would have come up with the two remaining solutions, hence the period of $\dfrac{\pi}{2}$. $$\tan(2x)=\sqrt{3}$$ $$2x=\dfrac{\pi}{3}+k\pi$$ $$x=\dfrac{\pi}{6}+k\dfrac{\pi}{2},\quad k\in\mathbb{Z}$$ 
March 29th, 2016, 07:00 PM  #4  
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2215  Quote:
As the cycle length is a multiple of the period for this question, you can alternatively add the cycle length, $\pi$, to each of your initial solutions to get the solutions in the second cycle. To avoid confusion, I recommend that you start by finding the solution(s) in the first period (rather than cycle) and then add the period length repeatedly until the specified domain is exhausted.  

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