My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum


Thanks Tree1Thanks
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
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.
davedave is offline  
 
April 13th, 2016, 01:01 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 20,475
Thanks: 2039

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
skipjack is offline  
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.
davedave is offline  
April 13th, 2016, 02:45 PM   #4
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,885
Thanks: 1504

Quote:
Originally Posted by davedave View Post
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 ...

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

$\cos^2{x} = \dfrac{3}{4} \implies \cos{x} = \pm \dfrac{\sqrt{3}}{2} \implies x = \dfrac{\pi}{6} \, , \, \dfrac{5\pi}{6} \, , \, \dfrac{7\pi}{6}\, , \, \dfrac{11\pi}{6}$
skeeter is online now  
April 14th, 2016, 11:24 PM   #5
Senior Member
 
Joined: Apr 2008

Posts: 194
Thanks: 3

Thanks a lot, skeeter.
davedave is offline  
April 15th, 2016, 01:14 AM   #6
Global Moderator
 
Joined: Dec 2006

Posts: 20,475
Thanks: 2039

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 $11\pi/6$.
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Trigonometry

Tags
find, miss, missing, solutions



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Find the missing number 10 - 6 * 15 / ? = 20 CountryThrills Algebra 2 November 17th, 2015 11:33 AM
Puzzle. Find the missing number. snape Elementary Math 7 February 4th, 2013 11:28 PM
Find the missing base 100101 two = 31 ______ vivalajuicy Number Theory 1 November 11th, 2012 02:06 PM
Find the missing number zgonda New Users 0 August 29th, 2010 05:46 PM
Find missing constants in polynomial limit unwisetome3 Algebra 0 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.