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    Again with the last thread 0°≤x≤360°

    4(sin²x-cosx)=3-2cosx

    Here's what I've done

    4(1-cos²x-cosx)=3-2cosx
    4-4cos²x-4cosx=3-2cosx
    1-4cos²x-2cosx=0
    4cos²x+2cosx=1
    2cosx(2cosx+1)=1


    2cosx=1
    where cosx=0.5
    x= 60 and 300 (Using CAST diagram)

    2cosx+1=1
    where 2cosx=0
    x=90 and 270 (using graph of cos)

    where did i go wrong?
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    (Original post by thefatone)
    Again with the last thread 0°≤x≤360°

    4(sin²x-cosx)=3-2cosx

    Here's what I've done

    4(1-cos²x-cosx)=3-2cosx
    4-4cos²x-4cosx=3-2cosx
    1-4cos²x-2cosx=0
    4cos²x+2cosx=1
    2cosx(2cosx+1)=1


    2cosx=1
    where cosx=0.5
    x= 60 and 300 (Using CAST diagram)

    2cosx+1=1
    where 2cosx=0
    x=90 and 270 (using graph of cos)

    where did i go wrong?
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    (Original post by thefatone)
    2cosx(2cosx+1)=1


    2cosx=1
    You can only do that if it's =0, you have =1, make it a quadratic in cos x =0 and then solve.
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    Remember that  \arccos(\frac{\sqrt5-1}{4}) = 72^{\circ} .
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    (Original post by Zacken)
    You can only do that if it's =0, you have =1, make it a quadratic in cos x =0 and then solve.
    ah okay thanks
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    (Original post by TeeEm)
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    lol just realised xD
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    (Original post by thefatone)
    lol just realised xD
    no worries ...
    (hopefully you liked the picture ...)
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    (Original post by TeeEm)
    no worries ...
    (hopefully you liked the picture ...)
    i laughed for a good 10 seconds when i saw the highlighted red bits and the picture xD
    still a bit giddy now
 
 
 

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