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Math Help - Cos2x?`

  1. #1
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    Cos2x?`

    Find the solutions to the following equations in the interval [0,2 ].
    (c) sin2x=−23 .

    Note.
    sin2x means sin(2x) and not (sin2)x.
    Hint. The angle
    x belongs to the interval [0,2 ] if and only if the angle 2x belongs to the interval [0,4 ]. So, how many solutions does the given equation have in the interval [0,2 ] ?

    What is this asking? I got 4pi/3 and 5p/3 but it's obviously not right.
    Also the title should obviously be sin 2x not cos..
    Last edited by katekate; September 21st 2009 at 07:45 PM. Reason: Title Error
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  2. #2
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    Hi katekate

    Are you sure the question is \sin 2x = -2 \sqrt {3} , not \sin 2x = -\frac{1}{2} \sqrt {3} ?

    If it is, then no value of x can satisfy that equation.
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  3. #3
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    oh wow sorry i copied and pasted it from something and it obviously didn't transfer. Yeah it should be sqrt(3)/-2
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  4. #4
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    Quote Originally Posted by katekate View Post
    oh wow sorry i copied and pasted it from something and it obviously didn't transfer. Yeah it should be sqrt(3)/-2

    \sin 2x=-\frac{\sqrt{3}}{2}

    2x=\frac{4\pi}{3} , \frac{5\pi}{3} , \frac{10\pi}{3} , \frac{11\pi}{3}

    Then x=\frac{2\pi}{3} , \frac{5\pi}{6} , \frac{5\pi}{3} , \frac{11\pi}{6}

    Few things to note

    (1) 0<2x<4\pi

    (2) \sin 60 = \frac{\sqrt{3}}{2} , special angle .
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  5. #5
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    Hi katekate

    Addition to what mathaddict had explained :

    \sin 2x=-\frac{\sqrt{3}}{2}

    \sin 2x = \sin \left(-\frac{\pi}{3}\right)

    2x=-\frac{\pi}{3}

    x=-\frac{\pi}{6}

    Because sin is periodic with period 2n\pi ,where n is integer, the value of sin (2x) will be the same as \sin (2x \pm 2n\pi). Then :
    1. for n = 1
    sin (2x) = sin (2x - 2\pi). So : 2x - 2\pi = -\frac{\pi}{3} ----> find x

    2. for n = 2
    sin (2x) = sin (2x - 4\pi). So ....

    Continue until finding all values of x that satisfy the equation



    Or you can use formula (general solution for sin) :

    \sin \theta = sin \alpha

    \theta = (-1)^n*\alpha + n\pi , where n is integer

    So:

    \sin 2x = -\frac{\sqrt{3}}{2}

    2x=(-1)^n* \left(-\frac{\pi}{3}\right)+n\pi

    x=(-1)^n* \left(-\frac{\pi}{6}\right)+\frac{1}{2}n\pi

    1) for n = 0

    x=-\frac{\pi}{6} ----> we reject this

    2) for n = 1

    ......

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