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 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 ${\le}$x<$2{\pi}$ This is how I do it. $\frac{2\tan(x)}{1-(\tan(x))^2}$ = $\frac{1}{\tan(x)}$ After doing some algebra, I get the equation below. $\(\tan(x))^2$ = $\frac{1}{3}$ Next, I square root both sides and get two equations. tan(x) = $\frac{1}{sqrt3}$ and tan(x) = - $\frac{1}{sqrt3}$ The reference angle is $\frac{\pi}{6}$. For the equation with the positive root, I find the two solutions in quadrants I and III which are x = $\frac{\pi}{6}$ and $\frac{7\pi}{6}$. For the equation with the negative root, I find the two solutions in quadrants II and IV which are x = $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$. In addition to the four solutions above, the answer key also gives two more which are x = $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. 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
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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 $11\pi/6$.

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