Trig question

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  1. HEY_101's Avatar
    1 05-05-2012  09:40
    • Adored and Respected Member
    • Location: London
    • Posts: 461
    Trig question
    Hey guys,

    I was wondering if you could explain this question to me
    solve the equation cos2x= the squarerootof 3 sin2x

    if i bring sin 2x under cos2x to get tan2x

    so the new equation would be tan2x= squareroot of 3 would it not?
  2. raheem94's Avatar
    2 05-05-2012  09:48
    • TSR Demigod
    • Posts: 5,512
    Re: Trig question
    (Original post by HEY_101)
    Hey guys,

    I was wondering if you could explain this question to me
    solve the equation cos2x= the squarerootof 3 sin2x

    if i bring sin 2x under cos2x to get tan2x

    so the new equation would be tan2x= squareroot of 3 would it not?
    \displaystyle cos2x = \sqrt{3sin2x}

    Square both sides, to get, cos^22x=3sin2x

    Remember, cos^22x = 1-sin^22x

    This will give you a quadratic.
  3. TenOfThem's Avatar
    3 05-05-2012  09:52
    • TSR Royalty
    Re: Trig question
    Unless

    cos2x = \sqrt3 \sin2x

    in which case

    No

    \dfrac{cosx}{sinx} \not= tanx
  4. HEY_101's Avatar
    4 05-05-2012  11:11
    • Adored and Respected Member
    • Location: London
    • Posts: 461
    Re: Trig question
    (Original post by raheem94)
    \displaystyle cos2x = \sqrt{3sin2x}
    the square root is only on the 3 not the sin 2x
    Post rating:
    1
  5. raheem94's Avatar
    5 05-05-2012  15:51
    • TSR Demigod
    • Posts: 5,512
    Re: Trig question
    (Original post by HEY_101)
    I was thinking that as well but the answers says tan2x= 1/ square root of 3.
    \displaystyle cos2x = \sqrt3 sin2x

    Divide both sides by cos2x
    \displaystyle \frac{cos2x}{cos2x} = \frac{\sqrt3 sin2x}{cos2x} \implies 1 = \sqrt3 tan2x \implies tan2x = \frac{1}{\sqrt3}

    I don't understand why are you getting confused at such a simple thing.
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