User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 April 12th, 2016, 10:11 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 How can I find the two missing solutions? Solve tan(2x) = cot(x) for 0 x< This is how I do it. = After doing some algebra, I get the equation below. = Next, I square root both sides and get two equations. tan(x) = and tan(x) = - The reference angle is . For the equation with the positive root, I find the two solutions in quadrants I and III which are x = and . For the equation with the negative root, I find the two solutions in quadrants II and IV which are x = and . In addition to the four solutions above, the answer key also gives two more which are x = and . I am not sure how to find the last two solutions. Can someone explain it? Thank you very much. Last edited by skipjack; April 13th, 2016 at 12:49 AM. April 13th, 2016, 01:01 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 tan(x) isn't defined for x = $\pi$/2 or 3$\pi$/2, but cot(x) is. tan(2x) = cot(x) = tan(pi/2 - x) 2x = $\pi$/2 - x + k$\pi$, where k is an integer 3x = $\pi$/2 + k$\pi$ x = $\pi$/6 + k$\pi$/3 Now one chooses the values of k such that 0 $\small\leqslant$ x < 2$\pi$: k = 0, 1, 2, 3, 4, or 5. x = $\pi$/6, $\pi$/2, 5$\pi$/6, 7$\pi$/6, 3$\pi$/2, or 11$\pi$/6. Thanks from davedave April 13th, 2016, 02:01 PM #3 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 Thanks, skipjack. I would have never thought of approaching the problem the way you did. Your way is really a genius method. Is there a way I can find the two missing solutions by using my method? Thanks. April 13th, 2016, 02:45 PM   #4
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,017
Thanks: 1603

Quote:
 Originally Posted by davedave Is there a way I can find the two missing solutions by using my method?
change the original equation to sine and cosine ...

$\tan(2x) = \cot{x}$

$\dfrac{\sin(2x)}{\cos(2x)} - \dfrac{\cos{x}}{\sin{x}} = 0$

$\dfrac{2\sin{x}\cos{x}}{\cos^2{x}-\sin^2{x}} - \dfrac{\cos{x}}{\sin{x}} = 0$

$\dfrac{2\sin^2{x}\cos{x}-\cos{x}(\cos^2{x}-\sin^2{x})}{\sin{x}(\cos^2{x}-\sin^2{x})} = 0$

$\dfrac{3\sin^2{x}\cos{x}-\cos^3{x}}{\sin{x}(\cos^2{x}-\sin^2{x})} = 0$

$\dfrac{3(1-\cos^2{x})\cos{x}-\cos^3{x}}{\sin{x}(\cos^2{x}-\sin^2{x})} = 0$

$\dfrac{\cos{x}(3-4\cos^2{x})}{\sin{x}(\cos^2{x}-\sin^2{x})} = 0$

setting the numerator = 0 ...

$\cos{x} = 0 \implies x = \dfrac{\pi}{2} \, , \, \dfrac{3\pi}{2}$

$\cos^2{x} = \dfrac{3}{4} \implies \cos{x} = \pm \dfrac{\sqrt{3}}{2} \implies x = \dfrac{\pi}{6} \, , \, \dfrac{5\pi}{6} \, , \, \dfrac{7\pi}{6}\, , \, \dfrac{11\pi}{6}$ April 14th, 2016, 11:24 PM #5 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 Thanks a lot, skeeter. April 15th, 2016, 01:14 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 At your first step, you assumed that tan(x) is defined. The alternative is that x = $\pi/2$ or $3\pi/2$, which, by inspection, are solutions. Initially expanding tan(2x) in terms of cot(x) gives 2cot(x)/(cot²(x) - 1) = cot(x), which leads to cot(x)(cot²(x) - 3) = 0. Hence cot(x) = 0, which implies x = $\pi/2$ or $3\pi/2$, or cot(x) = ±√3, which implies x = $\pi/6$, $5\pi/6$, $7\pi/6$ or $11\pi/6$. Tags find, miss, missing, 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 CountryThrills Algebra 2 November 17th, 2015 11:33 AM snape Elementary Math 7 February 4th, 2013 11:28 PM vivalajuicy Number Theory 1 November 11th, 2012 02:06 PM zgonda New Users 0 August 29th, 2010 05:46 PM unwisetome3 Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      