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Homework Help: Another Trig Equation

  1. Oct 6, 2005 #1
    [tex]
    \begin{align*}
    \sin^2 \theta + 2 \sin \theta \cos \theta - 3 \cos^2 \theta = 0 \\
    (\sin \theta + 3 \cos \theta)(\sin \theta - \cos \theta) = 0 \\
    \end{align*}
    [/tex]

    so,

    [tex]
    \begin{align*}
    \sin \theta = \cos \theta\\
    \theta = \frac{\pi}{4} + \pi k, k \epsilon \mathbb{I}
    \end{align*}
    [/tex]

    or

    [tex]
    \begin{align*}
    \sin \theta = -3 \cos \theta\\
    \tan \theta = -3\\
    \theta = 5.03 + 2\pi k, k \epsilon \mathbb{I}; 1.89 + 2\pi k, k \epsilon \mathbb{I}
    \end{align*}
    [/tex]

    How come this is incorrect?

    ...argh I can't align it properly :uhh:
     
    Last edited: Oct 6, 2005
  2. jcsd
  3. Oct 6, 2005 #2

    hotvette

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    Homework Helper

    a x b = 0 doesn't mean both a and b are zero, although they could be. It simply means at least one of them is zero for the product to be zero.
     
  4. Oct 6, 2005 #3

    TD

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    Homework Helper

    Here, you can get all solution by taking pi/4 + k*pi instead of 2k*pi.
     
  5. Oct 6, 2005 #4
    Oops, I had that written down but I made the mistake when posting!

    Does this mean you say pi/4 is a correct answer? When I plug it back into the original equation I don't get 0 :frown:
     
  6. Oct 6, 2005 #5

    TD

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    Homework Helper

    How's that?

    [tex]
    \sin ^2 \theta + 2\sin \theta \cos \theta - 3\cos ^2 \theta \mathop \to \limits^{\theta = \pi /4} \left( {\sin \frac{\pi }{4}} \right)^2 + 2\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 3\left( {\cos \frac{\pi }{4}} \right)^2 = \frac{1}{2} + 2 \cdot \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - 3 \cdot \frac{1}{2} = 0
    [/tex]
     
  7. Oct 6, 2005 #6
    OK, I'm just bad at typing things into my calculator then! That's what I get for using it in the first place o:)

    By the way, can anyone tell me how to align things properly with tex? For some reason using "\\" won't skip lines for me.

    Thanks.
     
    Last edited: Oct 6, 2005
  8. Oct 6, 2005 #7

    TD

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    Homework Helper

    It will if you use an array:

    [tex]\begin{array}{l}
    x^2 - 4 = 0 \\
    x = 2 \vee x = - 2 \\
    \end{array}[/tex]

    Code (Text):

    \begin{array}{l}
     x^2  - 4 = 0 \\
     x = 2 \vee x =  - 2 \\
     \end{array}
     
     
  9. Oct 6, 2005 #8
    Doh! Thanks again!
     
  10. Oct 6, 2005 #9

    TD

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    Homework Helper

    You're welcome :smile:
     
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