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

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CallistoBest ResponseYou've already chosen the best response.2
Hmm... \[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
I did that and after that I got: \[\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2\]
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
2cosθsinθ = sin2θ Divide both sides by 2 cosθsinθ = ...?
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
\[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Multiply both sides by sin2θ, what do you get?
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
wait, haha I'm sorry!...
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Next: Use cos2x = 12(sinx)^2
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]
 one year ago

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

CallistoBest ResponseYou've already chosen the best response.2
It should be\[\sin^{1}\frac{ \sqrt{2} }{ 2 }=2\theta\]
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
\[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
now, how do I find the the equation in the interval [0, 2π)??
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
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 \[sin2\theta = \sqrt{\frac{1}{2}}\]
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
i noticed something, since the square root was taken to take out the squared sine....then the result needs to be + \[\pm \frac{ \sqrt{2} }{ 2 }\]
 one year ago

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

CallistoBest ResponseYou've already chosen the best response.2
But it's up to you :)
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
ok, so when I solve for all the solutions...is it?: \[\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }\]
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
From zero to pi, yes... keep going to 2pi... 4 more solutions.
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
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)
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
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
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
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. :)
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
I don't like to just give answers, but you are working very hard at this. ;)
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
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 :(
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
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.
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
ok... I stand by my answer... and @Callisto 's
 one year ago

Dama28Best ResponseYou've already chosen the best response.1
ok. hahahah thank you so sooooo much! and I'm sorry for the trouble
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
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.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Agree with @EulerGroupie 's saying :)
 one year ago
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