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Math Help - Solve the double angle equation

  1. #1
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    Solve the double angle equation

    Im stuck on this question in my maths homework:

    Solve the equation √3tan2x-1=0 for 0<X<π

    Here is what i've done:

    √3tan2x-1=0
    tan2x=1/√3
    Using trig identities,
    tanx= sinx/cosx
    so tan2x=sin2x/cos2x < is this right?
    I've then used the trig expansions:
    2sinxcosx/1-2sin^2x < Am I using the right expansion for cos2x?

    Am I using the right method? I have tried to go further than this, but nothing has worked out, so where do I go from here?

    Thanks in advance.
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  2. #2
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    Don't bother changing it to sin and cos - arctan is a fine operation and you can use it on your second line

    \arctan(\tan(2x)) = \arctan\left({\dfrac{1}{\sqrt{3}}\right) \implies 2x = \dfrac{\pi}{6} \implies x = \dfrac{\pi}{12}

    That is the principle solution but remember that since you have a double angle the period is \dfrac{\pi}{2} so you'll need to add that to get the second solution
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  3. #3
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    I would do that, but we havent been taught it in class and its not in our notes, so I have to do it the sin and cos way. Thanks anyway.
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  4. #4
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    Quote Originally Posted by Fionafaye View Post
    I would do that, but we havent been taught it in class and its not in our notes, so I have to do it the sin and cos way.
    Draw an equilateral triangle with sides of length two.
    Draw an altitude from the summit angle.
    It is also a median and angular bisector.
    The altitude has length \sqrt3 and the adjunct side has length 1.
    So half the summit angle is \frac{\pi}{6} and \tan\left(\frac{\pi}{6}\right)= \left(\frac{1}{\sqrt3}\right) .
    Now finish.
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  5. #5
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    If you want it in sin and cos, then

    √3sin(2x) = cos(2x)

    sin^2(x) + 2√3sin(x)cos(x) - cos^2(x) = 0

    Solve the quadratic for sin(x). From that find tan(x)
    Last edited by sa-ri-ga-ma; February 27th 2011 at 05:59 AM.
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  6. #6
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    a tricky kind of question.
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