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BloomLocke367

  • one year ago

Solve this in interval form, from 0 to 2pi. Round to the nearest hundredth. \(-3sin2\theta=1.5\)

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  1. BloomLocke367
    • one year ago
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    @mathmate @pooja195

  2. BloomLocke367
    • one year ago
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    @dan815 Will you come help?

  3. BloomLocke367
    • one year ago
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    can you help?

  4. BloomLocke367
    • one year ago
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    @pooja195

  5. BloomLocke367
    • one year ago
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    @ganeshie8 Can you please help?

  6. ganeshie8
    • one year ago
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    as a start, divide -3 both sides

  7. ganeshie8
    • one year ago
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    \[-3\sin(2\theta) = 1.5 \implies \sin(2\theta) = \frac{1.5}{-3} =-\frac{1}{2} \]

  8. ganeshie8
    • one year ago
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    \[\sin(2\theta) = -\frac{1}{2} \implies 2\theta = \sin^{-1}\left(-\frac{1}{2}\right)\] look up unit circle

  9. BloomLocke367
    • one year ago
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    I have the unit circle

  10. BloomLocke367
    • one year ago
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    what am I looking for?

  11. ganeshie8
    • one year ago
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    see for what angles the sin value is \(-\frac{1}{2}\)

  12. BloomLocke367
    • one year ago
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    sin is the y-value, right?

  13. ganeshie8
    • one year ago
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    |dw:1432912910233:dw|

  14. BloomLocke367
    • one year ago
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    so, 7pi/6 and 11pi/6

  15. BloomLocke367
    • one year ago
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    and I just round that to the nearest hundredth and I'll have my answer?

  16. ganeshie8
    • one year ago
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    Yes, \[2\theta = \frac{7\pi}{6}+2k\pi,~ ~\frac{11\pi}{6}+2k\pi\] that means \[\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi\]

  17. ganeshie8
    • one year ago
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    but you want solutions in the interval 0 to 2pi, right ?

  18. BloomLocke367
    • one year ago
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    yes

  19. ganeshie8
    • one year ago
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    plugin k = 0, 1, 2 etc till you hit 2pi

  20. BloomLocke367
    • one year ago
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    why?

  21. ganeshie8
    • one year ago
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    \[\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi\] plugin \(k=0\) and get \[\theta = \frac{7\pi}{12},~ ~\frac{11\pi}{12}\] plugin \(k=1\) and get \[\theta = \frac{7\pi}{12}+\pi,~ ~\frac{11\pi}{12}+\pi\]

  22. ganeshie8
    • one year ago
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    because any integer value of \(k\) gives an angle that is a solution to the given equation

  23. ganeshie8
    • one year ago
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    there are infinitely many solutions as there are infinitely many integers your question is about finding the solutions in the interval 0 to 2pi

  24. BloomLocke367
    • one year ago
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    okay.

  25. BloomLocke367
    • one year ago
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    and you plug in k=2, and get \(\large\frac{7\pi}{12}+2\pi,~\frac{11\pi}{12}+2\pi\)

  26. ganeshie8
    • one year ago
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    yes clearly thats more than \(2\pi\), so we may stop now.

  27. ganeshie8
    • one year ago
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    k=0 and k=1 are enough

  28. ganeshie8
    • one year ago
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    find the corresponding \(\theta\) values as you can see there are exactly four solutions in the interval 0 to 2pi

  29. BloomLocke367
    • one year ago
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    okay, and I just round it?

  30. ganeshie8
    • one year ago
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    the question asks you to round it so yeah

  31. BloomLocke367
    • one year ago
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    so, 1.83, 2.88, 4.97, and 6.02

  32. ganeshie8
    • one year ago
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    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

  33. BloomLocke367
    • one year ago
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    Okay, thanks!

  34. ganeshie8
    • one year ago
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    yw

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