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April 12th, 2016, 11:11 PM  #1 
Senior Member Joined: Apr 2008 Posts: 193 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 01:49 AM. 
April 13th, 2016, 02:01 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,270 Thanks: 1958 
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. 
April 13th, 2016, 03:01 PM  #3 
Senior Member Joined: Apr 2008 Posts: 193 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, 03:45 PM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,815 Thanks: 1458  Quote:
$\tan(2x) = \cot{x}$ $\dfrac{\sin(2x)}{\cos(2x)}  \dfrac{\cos{x}}{\sin{x}} = 0$ $\dfrac{2\sin{x}\cos{x}}{\cos^2{x}\sin^2{x}}  \dfrac{\cos{x}}{\sin{x}} = 0$ $\dfrac{2\sin^2{x}\cos{x}\cos{x}(\cos^2{x}\sin^2{x})}{\sin{x}(\cos^2{x}\sin^2{x})} = 0$ $\dfrac{3\sin^2{x}\cos{x}\cos^3{x}}{\sin{x}(\cos^2{x}\sin^2{x})} = 0$ $\dfrac{3(1\cos^2{x})\cos{x}\cos^3{x}}{\sin{x}(\cos^2{x}\sin^2{x})} = 0$ $\dfrac{\cos{x}(34\cos^2{x})}{\sin{x}(\cos^2{x}\sin^2{x})} = 0$ 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 15th, 2016, 12:24 AM  #5 
Senior Member Joined: Apr 2008 Posts: 193 Thanks: 3 
Thanks a lot, skeeter.

April 15th, 2016, 02:14 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,270 Thanks: 1958 
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 $11\pi/6$. 

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