Question

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π) :
tan θ + cot θ = 4 sin 2θ

Answer 1

Hmm...
\[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]

Answer 2

I did that and after that I got:
\[\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2\]

Answer 3

after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]

Answer 4

2cosθsinθ = sin2θ
Divide both sides by 2
cosθsinθ = ...?

Answer 5

\[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]

Answer 6

Multiply both sides by sin2θ, what do you get?

Answer 7

wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]

Answer 8

The first one.

Answer 9

wait, haha I'm sorry!...

Answer 10

its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]

Answer 11

Next:
Use
cos2x = 1-2(sinx)^2

Answer 12

so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]

Answer 13

riiiight! I see it.

Answer 14

\[cos2x = 1-2sin^2x\]\[2sin^2x = 1-cos2x\]Replace x by 2θ. Life isn't difficult then.

Answer 15

It should be\[\sin^{-1}\frac{ \sqrt{2} }{ 2 }=2\theta\]

Answer 16

i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]

Answer 17

Are you sure...???

Answer 18

\[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

Answer 19

2theta =pi/4, 3pi/4

Answer 20

so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??

Answer 21

yupp

Answer 22

now, how do I find the the equation in the interval [0, 2π)??

Answer 23

Hmm... Something's wrong there.
\[1=2\sin ^{2}2\theta\]\[sin^2 2\theta = \frac{1}{2}\]\[sin2\theta = \pm \sqrt{\frac{1}{2}}\]So, \[sin2\theta = \sqrt{\frac{1}{2}}\]or \[sin2\theta = -\sqrt{\frac{1}{2}}\]

Answer 24

i noticed something, since the square root was taken to take out the squared sine....then the result needs to be +- \[\pm \frac{ \sqrt{2} }{ 2 }\]

Answer 25

:)

Answer 26

wow! at the same time

Answer 27

To avoid taking square root, I prefer using my method:
\[1=2\sin ^{2}2\theta\]\[1=1-cos4\theta\]\[cos4\theta =0\]Then. solve theta.

Answer 28

ohhhh ok

Answer 29

But it's up to you :)

Answer 30

ok, so when I solve for all the solutions...is it?:

\[\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }\]

Answer 31

From zero to pi, yes... keep going to 2pi... 4 more solutions.

Answer 32

The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.

Answer 33

If you want to get the next 4 solutions, use: 2pi - solution you've got. Then you'll have 4 more.
Eg: 2pi - (pi/8)

Answer 34

What have you put?

Answer 35

pi/8, 3pi/8 my first try
and pi/8, 3pi/8, 5pi/8, 7pi/8 the second try. I have one more try :S

Answer 36

pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8 are all of the solutions between zero and 2pi. I wish I could show you my pretty graph on my calculator. :)

Answer 37

I don't like to just give answers, but you are working very hard at this. ;)

Answer 38

First four: pi/8, 3pi/8, 5pi/8, 7pi/8
Next four:
2pi - pi/8 = 15pi/8
2pi - 3pi/8 = 13pi/8
2pi - 5pi/8 =...
2pi - 7pi/8 =...

I'm sorry :(

Answer 39

My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.

Answer 40

hahaha thanks. the reason why I didn't keep going is because in the previous problems I had, if it was divided by 3 theta I would only give 6 answers. If if was divided by 4theta, then 8 answers, so I would assume that in this excercise that's divided by 2theta I would only give 4 answers.

Answer 41

its [0, 2pi)

Answer 42

ok... I stand by my answer... and @Callisto 's

Answer 43

ok. hahahah thank you so sooooo much! and I'm sorry for the trouble

Answer 44

Oh, this stuff is fun... no troubles. We do it because we love it... especially when people are really working and trying instead of fishing for easy answers.

Answer 45

ok. thanks!

Answer 46

Agree with @EulerGroupie 's saying :)