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Math Help - general solution ,trig?

  1. #1
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    general solution ,trig?

    What is the general solution to: 2sin2\Theta=\sqrt{3}tan2\Theta? any ideas?
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  2. #2
    Senior Member Spec's Avatar
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    2\sin 2\theta = \sqrt{3}\tan 2\theta = \sqrt{3}\frac{\sin 2\theta}{\cos 2\theta}\Longleftrightarrow \sin 2\theta = 0 or \cos 2\theta = \frac{\sqrt{3}}{2}
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  3. #3
    MHF Contributor Amer's Avatar
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    ok

    2sin2\theta=\sqrt{3}tan2\theta


    2sin2\theta=\sqrt{3}\left(\frac{sin2\theta}{cos2\t  heta}\right)

    dived the to side sin2 theta

    2=\sqrt{3}\left(\frac{1}{cos2\theta}\right)

    cos2\theta=\frac{\sqrt{3}}{2}

    2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\  pi
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  4. #4
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    would the end result be \frac{\pi}{12}+n\pi or would it be \frac{\pi}{12}+2n\pi?
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  5. #5
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    You can't just divide away \sin 2\theta from both sides since it's not a constant. If you do that, you'll miss out on all the solutions where \sin 2\theta=0

    It's similar to this equation: x^3=x has obviously got one root x=0, but if you divide both sides with x without taking that into consideration you'll end up with x^2=1, and thus miss a root.

    The general solution to \sin x = C is x=v+2n\pi or x=\pi-v+2n\pi where v is the angle and n\in Z

    The general solution to \cos x = C is x=\pm v+2n\pi where v is the angle and n\in Z
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  6. #6
    Super Member dhiab's Avatar
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    resolution

    good night, I offer you a resolution
    2shared - download R1.pdf
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  7. #7
    Super Member dhiab's Avatar
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    Quote Originally Posted by Amer View Post
    ok

    2sin2\theta=\sqrt{3}tan2\theta


    2sin2\theta=\sqrt{3}\left(\frac{sin2\theta}{cos2\t  heta}\right)

    dived the to side sin2 theta

    2=\sqrt{3}\left(\frac{1}{cos2\theta}\right)

    cos2\theta=\frac{\sqrt{3}}{2}

    2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\  pi
    good night, you not to pouver not to divide on sin 2theta
    because sin2 theta can etre egal in zero
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  8. #8
    MHF Contributor Amer's Avatar
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    I am sorry I should be careful when I answer questions
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