 My Math Forum Why can't I add the period to find the next two solutions?
 User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 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:
 Originally Posted by davedave Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle.
You cover two periods in each cycle, so adding the period to your first solution, $\displaystyle \frac{\pi}{6}$, gives your second solution in the same cycle, $\displaystyle \frac{2\pi}{3}$. Adding the period to your second solution gives the first solution in the second cycle, $\displaystyle \frac{7\pi}{6}$. Adding the period once more will then give the second solution in the second cycle, $\displaystyle \frac{5\pi}{3}$.

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. Tags add, find, period, solutions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post X21 Trigonometry 4 January 4th, 2016 06:50 PM Albert.Teng Algebra 6 December 28th, 2013 06:44 AM n777l Algebra 2 September 20th, 2013 04:56 PM ajaxanon Algebra 6 June 1st, 2010 08:02 PM 450081592 Linear Algebra 3 November 29th, 2009 10:20 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      