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anonymous
 3 years ago
Use a Double or HalfAngle Formula to solve the equation in the interval [0, 2π) :
tan θ + cot θ = 4 sin 2θ
anonymous
 3 years ago
Use a Double or HalfAngle Formula to solve the equation in the interval [0, 2π) : tan θ + cot θ = 4 sin 2θ

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Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2Hmm... \[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did that and after that I got: \[\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.22cosθsinθ = sin2θ Divide both sides by 2 cosθsinθ = ...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2Multiply both sides by sin2θ, what do you get?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, haha I'm sorry!...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2Next: Use cos2x = 12(sinx)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2\[cos2x = 12sin^2x\]\[2sin^2x = 1cos2x\]Replace x by 2θ. Life isn't difficult then.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2It should be\[\sin^{1}\frac{ \sqrt{2} }{ 2 }=2\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2\[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now, how do I find the the equation in the interval [0, 2π)??

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2Hmm... 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
 3 years ago
Best ResponseYou've already chosen the best response.0i noticed something, since the square root was taken to take out the squared sine....then the result needs to be + \[\pm \frac{ \sqrt{2} }{ 2 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow! at the same time

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2To avoid taking square root, I prefer using my method: \[1=2\sin ^{2}2\theta\]\[1=1cos4\theta\]\[cos4\theta =0\]Then. solve theta.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, 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
 3 years ago
Best ResponseYou've already chosen the best response.0From zero to pi, yes... keep going to 2pi... 4 more solutions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2If you want to get the next 4 solutions, use: 2pi  solution you've got. Then you'll have 4 more. Eg: 2pi  (pi/8)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0pi/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
 3 years ago
Best ResponseYou've already chosen the best response.0pi/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
 3 years ago
Best ResponseYou've already chosen the best response.0I don't like to just give answers, but you are working very hard at this. ;)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2First 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
 3 years ago
Best ResponseYou've already chosen the best response.0My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hahaha 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok... I stand by my answer... and @Callisto 's

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok. hahahah thank you so sooooo much! and I'm sorry for the trouble

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.2Agree with @EulerGroupie 's saying :)
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