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Dama28
Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π) : tan θ + cot θ = 4 sin 2θ
Hmm... \[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]
I did that and after that I got: \[\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2\]
after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]
2cosθsinθ = sin2θ Divide both sides by 2 cosθsinθ = ...?
\[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]
Multiply both sides by sin2θ, what do you get?
wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]
wait, haha I'm sorry!...
its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]
Next: Use cos2x = 1-2(sinx)^2
so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]
\[cos2x = 1-2sin^2x\]\[2sin^2x = 1-cos2x\]Replace x by 2θ. Life isn't difficult then.
It should be\[\sin^{-1}\frac{ \sqrt{2} }{ 2 }=2\theta\]
i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]
\[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]
so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??
now, how do I find the the equation in the interval [0, 2π)??
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 \[sin2\theta = -\sqrt{\frac{1}{2}}\]
i noticed something, since the square root was taken to take out the squared sine....then the result needs to be +- \[\pm \frac{ \sqrt{2} }{ 2 }\]
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.
ok, so when I solve for all the solutions...is it?: \[\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }\]
From zero to pi, yes... keep going to 2pi... 4 more solutions.
The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.
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)
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
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. :)
I don't like to just give answers, but you are working very hard at this. ;)
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 :(
My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.
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.
ok... I stand by my answer... and @Callisto 's
ok. hahahah thank you so sooooo much! and I'm sorry for the trouble
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.
Agree with @EulerGroupie 's saying :)