BloomLocke367
  • BloomLocke367
Solve this in interval form, from 0 to 2pi. Round to the nearest hundredth. \(-3sin2\theta=1.5\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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BloomLocke367
  • BloomLocke367
@mathmate @pooja195
BloomLocke367
  • BloomLocke367
@dan815 Will you come help?
BloomLocke367
  • BloomLocke367
can you help?

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BloomLocke367
  • BloomLocke367
@pooja195
BloomLocke367
  • BloomLocke367
@ganeshie8 Can you please help?
ganeshie8
  • ganeshie8
as a start, divide -3 both sides
ganeshie8
  • ganeshie8
\[-3\sin(2\theta) = 1.5 \implies \sin(2\theta) = \frac{1.5}{-3} =-\frac{1}{2} \]
ganeshie8
  • ganeshie8
\[\sin(2\theta) = -\frac{1}{2} \implies 2\theta = \sin^{-1}\left(-\frac{1}{2}\right)\] look up unit circle
BloomLocke367
  • BloomLocke367
I have the unit circle
BloomLocke367
  • BloomLocke367
what am I looking for?
ganeshie8
  • ganeshie8
see for what angles the sin value is \(-\frac{1}{2}\)
BloomLocke367
  • BloomLocke367
sin is the y-value, right?
ganeshie8
  • ganeshie8
|dw:1432912910233:dw|
BloomLocke367
  • BloomLocke367
so, 7pi/6 and 11pi/6
BloomLocke367
  • BloomLocke367
and I just round that to the nearest hundredth and I'll have my answer?
ganeshie8
  • ganeshie8
Yes, \[2\theta = \frac{7\pi}{6}+2k\pi,~ ~\frac{11\pi}{6}+2k\pi\] that means \[\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi\]
ganeshie8
  • ganeshie8
but you want solutions in the interval 0 to 2pi, right ?
BloomLocke367
  • BloomLocke367
yes
ganeshie8
  • ganeshie8
plugin k = 0, 1, 2 etc till you hit 2pi
BloomLocke367
  • BloomLocke367
why?
ganeshie8
  • ganeshie8
\[\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi\] plugin \(k=0\) and get \[\theta = \frac{7\pi}{12},~ ~\frac{11\pi}{12}\] plugin \(k=1\) and get \[\theta = \frac{7\pi}{12}+\pi,~ ~\frac{11\pi}{12}+\pi\]
ganeshie8
  • ganeshie8
because any integer value of \(k\) gives an angle that is a solution to the given equation
ganeshie8
  • ganeshie8
there are infinitely many solutions as there are infinitely many integers your question is about finding the solutions in the interval 0 to 2pi
BloomLocke367
  • BloomLocke367
okay.
BloomLocke367
  • BloomLocke367
and you plug in k=2, and get \(\large\frac{7\pi}{12}+2\pi,~\frac{11\pi}{12}+2\pi\)
ganeshie8
  • ganeshie8
yes clearly thats more than \(2\pi\), so we may stop now.
ganeshie8
  • ganeshie8
k=0 and k=1 are enough
ganeshie8
  • ganeshie8
find the corresponding \(\theta\) values as you can see there are exactly four solutions in the interval 0 to 2pi
BloomLocke367
  • BloomLocke367
okay, and I just round it?
ganeshie8
  • ganeshie8
the question asks you to round it so yeah
BloomLocke367
  • BloomLocke367
so, 1.83, 2.88, 4.97, and 6.02
ganeshie8
  • ganeshie8
Looks good, you may double check with wolfram http://www.wolframalpha.com/input/?i=solve%20-3sin(2%5Ctheta)%3D1.5%2C%200%3C%5Ctheta%3C2pi&t=crmtb01
BloomLocke367
  • BloomLocke367
Okay, thanks!
ganeshie8
  • ganeshie8
yw

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