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Math Help - Intersections

  1. #1
    Senior Member
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    Intersections

    I can't find the intersections of (sqrt3)cos2x and sin2x. I know the basic method is to create a new equation from the two and obtain the x-intercepts.
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  2. #2
    MHF Contributor
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    Hi

    If I have well understood you need to solve
    \sqrt{3} \cos(2x) = \sin(2x)

    \sqrt{3} \cos(2x) - \sin(2x) = 0

    \frac{\sqrt{3}}{2} \cos(2x) - \frac12 \sin(2x) = 0

    \cos \left(\frac{\pi}{6}\right) \cos(2x) - \sin \left(\frac{\pi}{6}\right) \sin(2x) = 0

    \cos \left(2x + \frac{\pi}{6}\right) = 0

    Can you get it from here ?
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  3. #3
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    Hello, Stuck Man!

    Find the intersections of: . y \:=\:\sqrt{3}\cos2x\:\text{ and }\:y \:=\:\sin2x

    We have: . \sin2x \;=\;\sqrt{3}\cos2x

    Then: . \frac{\sin2x}{\cos2x} \:=\:\sqrt{3} \quad\Rightarrow\quad \tan2x \:=\:\sqrt{3}

    Hence: . 2x \;=\;\frac{\pi}{3} + \pi n

    Therefore: . x \;=\;\frac{\pi}{6} + \frac{\pi}{2}n

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  4. #4
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    Yes, thanks.
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