Jamierox4ev3r
  • Jamierox4ev3r
Solve:
Mathematics
chestercat
  • chestercat
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Jamierox4ev3r
  • Jamierox4ev3r
Give me a second to type the equation, thanks :)
Jamierox4ev3r
  • Jamierox4ev3r
Solve: \[2\tan ^{2}\theta \sin \theta - \tan ^{2}\theta=0\] for all values in [0, 360] <---degrees
Jamierox4ev3r
  • Jamierox4ev3r
@ganeshie8 any ideas?

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ganeshie8
  • ganeshie8
it seems \(\tan^2\theta\) is common in both the terms you may start by factoring it out
Jamierox4ev3r
  • Jamierox4ev3r
Alright. So that would be \( \tan( 2\tan \theta \sin \theta-\tan \theta )\) right?
ganeshie8
  • ganeshie8
pull out the entire \(\tan^2\theta\)
Jamierox4ev3r
  • Jamierox4ev3r
the entire thing? then that would leave you with.. just sine I believe. So \(\tan^{2} \theta (2\sin \theta-\theta)\)
ganeshie8
  • ganeshie8
wait a sec, \(\tan^{2} \theta (2\sin \theta-\color{Red}{\theta})\) are you really sure that \(\color{Red}{\theta}\) stays back there ?
Jamierox4ev3r
  • Jamierox4ev3r
wait... no it wouldn't. whoops xD
ganeshie8
  • ganeshie8
\(\tan^2\theta\) is ONE SINGLE thing \(\tan^2\) and \(\theta\) are not two pieces
Jamierox4ev3r
  • Jamierox4ev3r
^ yes I recall that now. But wouldn't something have to stay there?
Jamierox4ev3r
  • Jamierox4ev3r
oh wait. wouldn't it just be -1?
ganeshie8
  • ganeshie8
good, please fix it also what happened to \(= 0\)
Jamierox4ev3r
  • Jamierox4ev3r
oh it's still there, I just lopped it off. So as of now, this is where we are. \(\tan^{2} \theta (2\sin \theta-1)=0\)
ganeshie8
  • ganeshie8
looks good! what are you going to do next
Jamierox4ev3r
  • Jamierox4ev3r
I'm not sure....seems to me like there is a trigonometric identity somewhere in there perhaps.
ganeshie8
  • ganeshie8
Easy, actually we're almost done! all we need to know is the "zero product property"
ganeshie8
  • ganeshie8
|dw:1441068917025:dw|
ganeshie8
  • ganeshie8
so far we have : \(\color{red}{\tan^{2} \theta} \color{blue}{(2\sin \theta-1)}=0\) By zero product property, \(\color{red}{\tan^2\theta}=0\) or \(\color{blue}{2\sin\theta-1}=0\) solve each of them separately using unit circle
Jamierox4ev3r
  • Jamierox4ev3r
so we could set either one to zero. And solve for theta \((\theta)\)
ganeshie8
  • ganeshie8
Exactly! thats what zero product property tells us
Jamierox4ev3r
  • Jamierox4ev3r
wow. that actually makes so much sense. Alright, so let me solve each separately and I'll be back with you on what I got
ganeshie8
  • ganeshie8
take ur time you may have to use unit circle as you're asked to find all the solutions between 0 and 360
Jamierox4ev3r
  • Jamierox4ev3r
Noted.
Jamierox4ev3r
  • Jamierox4ev3r
Alright. So I got that \(\Large\sin \theta = \frac{1}{2}\) So from that, I found that \(\Large\theta = 30 (degrees), 150 (degrees)\)
Jamierox4ev3r
  • Jamierox4ev3r
I wasn't as sure for \(\tan \theta\), but I think that \(\Large\tan \theta =0\) and from that, wouldn't \(\Large\theta = 0 (degrees)\) ?? **not sure**
ganeshie8
  • ganeshie8
Excellent! \(\tan\theta=0\) should give you 3 solutions
ganeshie8
  • ganeshie8
il give a hint : \(\tan\theta = \tan(\theta+180)\)
Jamierox4ev3r
  • Jamierox4ev3r
oh wait i think i see. 90 degrees would be undefined
ganeshie8
  • ganeshie8
In other words : adding 180 to the angle wont change the value of \(\tan(\theta)\)
Jamierox4ev3r
  • Jamierox4ev3r
right.
Jamierox4ev3r
  • Jamierox4ev3r
I get that.
ganeshie8
  • ganeshie8
\(\tan\theta=0\) \(\implies \theta = \{0,\pm 180,\pm 360,\pm540,\ldots\}\) since you want the solutions between 0 and 360 you just pick \(0,180,360\)
Jamierox4ev3r
  • Jamierox4ev3r
oh and btw, I can include 0 degrees, but not 360. I was supposed to write the problem like [0,360) instead of [0,360]. So technically, for this problem, there are only two solutions that come from \(\tan \theta =0\)
ganeshie8
  • ganeshie8
Oh then you're absolutely right!
Jamierox4ev3r
  • Jamierox4ev3r
alright, so in total, all the solutions are 0, 30, 150, and 180! (all in degrees) oh my goodness, thank you so much! This review material was so distant in my memory, until you came along of course. Your patience is outstanding, thank you
ganeshie8
  • ganeshie8
np your patience is outstanding too! keep it up :)
Jamierox4ev3r
  • Jamierox4ev3r
d'aw you flatter me :')

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