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

- anonymous

- chestercat

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- Callisto

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

- anonymous

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

- anonymous

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

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## More answers

- Callisto

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

- anonymous

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

- Callisto

Multiply both sides by sin2θ, what do you get?

- anonymous

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

- Callisto

The first one.

- anonymous

wait, haha I'm sorry!...

- anonymous

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

- Callisto

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

- anonymous

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

- anonymous

riiiight! I see it.

- Callisto

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

- Callisto

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

- anonymous

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

- Callisto

Are you sure...???

- Callisto

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

- anonymous

2theta =pi/4, 3pi/4

- anonymous

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

- anonymous

yupp

- anonymous

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

- Callisto

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 \[sin2\theta = -\sqrt{\frac{1}{2}}\]

- anonymous

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

- Callisto

:)

- anonymous

wow! at the same time

- Callisto

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.

- anonymous

ohhhh ok

- Callisto

But it's up to you :)

- anonymous

ok, so when I solve for all the solutions...is it?:
\[\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }\]

- anonymous

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

- anonymous

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

- Callisto

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)

- Callisto

What have you put?

- anonymous

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

- anonymous

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. :)

- anonymous

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

- Callisto

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 :(

- anonymous

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

- anonymous

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.

- anonymous

its [0, 2pi)

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

ok. thanks!

- Callisto

Agree with @EulerGroupie 's saying :)

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