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Math Help - Quadratic trig equation #2

  1. #1
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    Quadratic trig equation #2

    and

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

    I got down half of it 3x = arcsin(1/2)

    The other half is what is confusing me. How do I get the values for x on this one? 3x = arcsin(+/- sqrt(1))

    I know arcsin is opposite / hypotenuse.
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  2. #2
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    Quote Originally Posted by FaRmBoX View Post
    and

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

    I got down half of it 3x = arcsin(1/2)

    The other half is what is confusing me. How do I get the values for x on this one? 3x = arcsin(+/- sqrt(1))

    I know arcsin is opposite / hypotenuse.
    2\sin^3(3x) - \sin^2(3x) - 2\sin(3x) +1 = 0

    This time solve the cubic

    2a^3 - a^2 - 2a +1 = 0

    what values for a did you get?

    How did you go with the first question you posted?
    Last edited by mr fantastic; January 4th 2010 at 05:31 PM. Reason: Moved question (and its replies) to new thread
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  3. #3
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    I got a = +/- squareroot(1) and a = 1/2
    Last edited by mr fantastic; January 4th 2010 at 05:31 PM.
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  4. #4
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     3x = \arcsin(1/2)

     3x = \frac{\pi}{6}

     x = \frac{\pi}{18}

    Also

     \sin(3x) = \pm \sqrt{1}

     \sin(3x) = -1,1

    Can you take it from here?
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  5. #5
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    For  3x = \arcsin(1/2)

    I also got  3x = \frac{5\pi}{6}
     x = \frac{5\pi}{18}

    Is that right?

    For the bottom, did I understand this correctly?  \pm \sqrt{1} = -1,1
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  6. #6
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    Quote Originally Posted by FaRmBoX View Post
    For  3x = \arcsin(1/2)

    I also got  3x = \frac{5\pi}{6}
     x = \frac{5\pi}{18}

    Is that right?
    Yep, as sin has more than one solution for that value.

    The best solution will look a little like

     x = \frac{(4n+1)\pi}{18}, n=0,1,2\dots


    Quote Originally Posted by FaRmBoX View Post
    For [tex]
    For the bottom, did I understand this correctly?  \pm \sqrt{1} = -1,1
    Yep the square root of 1 is -1 and 1.
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