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    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?
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    (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.
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    Unless

    cos2x = \sqrt3 \sin2x

    in which case

    No

    \dfrac{cosx}{sinx} \not= tanx
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    (Original post by raheem94)
     \displaystyle cos2x = \sqrt{3sin2x}
    the square root is only on the 3 not the sin 2x
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    (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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