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Thread: Trigonometry

  1. August 31st 2008, 08:52 AM #1
    nystudent2729
    nystudent2729 is offline
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    Question Trigonometry

    Find all solutions on the interval 0 is less than or equal to x is less than 2π:
    cos²x-cos(2x)=0
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  2. August 31st 2008, 09:11 AM #2
    Chop Suey
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    \cos{A \pm B} = \cos{A}\cos{B} \mp \sin{A}\sin{B}

    \cos{2x} = \cos{(x + x)}
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  3. August 31st 2008, 09:12 AM #3
    Sean12345
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    Hi nystudent2729,

    We have \cos^2(x)-\cos(2x)=0 Note that cos(2x) \equiv 2\cos^2(x)-1 Thus we are left with

    \cos^2(x)-(2\cos^2(x)-1)=1-\cos^2(x)=0

    \implies \cos^2(x)=1 \implies \cos(x)=\pm1

    Can you continue?
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