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Math Help - [Urgent] Solving Trig Equations

  1. #1
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    [Urgent] Solving Trig Equations

    Solve each equation for 0 \leq \theta \leq 360

    3sinx = 2cos^{2}x

    The answers are:
    30 and 150 degrees

    Nevermind, solved it
    Last edited by Macleef; November 21st 2007 at 03:09 PM.
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  2. #2
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    Quote Originally Posted by Macleef View Post
    3sinx = 2cos^{2}x

    3sinx = 1 - 2sin^{2}x
    You ate a 2, it should be 3\sin x=2-2\sin^2x.

    It is a simple quadratic.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    You ate a 2, it should be 3\sin x=2-2\sin^2x.

    It is a simple quadratic.
    What? I don't get what you mean by "you ate a 2"?
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  4. #4
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    You forgot a 2.

    Now, can you solve the remaining problem?
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  5. #5
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    Can you solve 2z^2  + 3z - 1 = 0?
    Then let \sin (x) = z.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Can you solve 2z^2  + 3z - 1 = 0?
    Then let \sin (x) = z.

    You can't solve for that equation...but I can't say there aren't any solutions because there are and I don't know what I'm doing wrong
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    You forgot a 2.

    Now, can you solve the remaining problem?

    I still don't know what you're saying? I didn't forget any 2
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  8. #8
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    Quote Originally Posted by Macleef View Post
    You can't solve for that equation...but I can't say there aren't any solutions because there are and I don't know what I'm doing wrong
    Sorry to learn that. Have you considered that you may not be ready to do this problem?
    If I were you, I would ask myself why I am being asked to do this level of problem?
    Do you have the necessary mathematical background to tackle a problem at this level?
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