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Math Help - trig question!

  1. #1
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    Talking trig question!

    solve the equation:
    root3*sin2 \theta+2sin^2 \theta=1 for 0< \theta< \pi

    thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by snowball View Post
    solve the equation:
    root3*sin2 \theta+2sin^2 \theta=1 for 0< \theta< \pi

    thanks!
    I'll nudge you in the right direction:

    \sqrt{3}\sin(2\vartheta)+2\sin^2\vartheta=1

    Note that \cos(2\vartheta)=1-2\sin^2\vartheta.

    So we end up with \sqrt{3}\sin(2\vartheta)=1-2\sin^2\vartheta\implies \sqrt{3}\sin(2\vartheta)=\cos(2\vartheta)

    Thus we end up with \sqrt{3}=\cot(2\vartheta)

    Can you take it from here?

    --Chris
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  3. #3
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    Talking

    thanks,that's very helpful, i can take it from there
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