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Homework Help: Need help forgot how to do trig question

  1. Jun 27, 2011 #1
    I tried doing this problem but I don't think it is right can someone help me?
    2(sin(x))^2+3sin(x)=-1 over the interval [0,2pi)
     
  2. jcsd
  3. Jun 27, 2011 #2

    eumyang

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    Homework Helper

    Show us what you've tried, and then we might be able to help you.
     
  4. Jun 27, 2011 #3
    I tried,

    2(sin(x))^2+3sin(x)=-1

    2(sin(x))+3sin(x)+1=0

    x=(-3(+/-)sqr 9-4(2)(1))/2(2)

    x=(-3(+/-)1)/4

    sinx=(-3(+/-)1)/4

    x=arcsin(-3(+/-)1)/4

    x=-90 or -30

    x=-1.571 or -.524
     
  5. Jun 27, 2011 #4

    eumyang

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    There's a typo in the 2nd line, and the last two lines aren't technically correct. You need to put the "sin" in front of the x. Also, you really didn't need to use the quadratic formula. This expression on the LHS:
    [tex]2\sin^2 x + 3\sin x + 1 = 0[/tex]
    is factorable.

    Some problems here. First, these answers are not in the interval [0, 2pi). Just add 2pi to these answers and you'll be okay.

    Second, the range of the arcsin function is only [-pi/2, pi/2], so there are actually 3 solutions that I see, not 2. (There is another angle whose sin is -1/2.)
     
  6. Jun 27, 2011 #5
    The question asks you to find the solution in [0,2pi). The answer you report is not.
    For what value of x in [0,2pi) is sin(x) = -1 OR -1/2?
     
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