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Even moar trig Watch

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    Same interval for x as last time
    0°≤x≤360°

    4cosx(cosx-1)=-5cosx

    i did
    4cos²x-4cosx=-5cosx
    4cos²x+cosx=0
    cosx(4cosx+1)=0

    but there's no answer in the back of the book which says that 0°,180°or 360° are answers
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    (Original post by thefatone)
    Same interval for x as last time
    0°≤x≤360°

    4cosx(cosx-1)=-5cosx

    i did
    4cos²x-4cosx=-5cosx
    4cos²x+cosx=0
    cosx(4cosx+1)=0

    but there's no answer in the back of the book which says that 0°,180°or 360° are answers
    That would be because \cos x = 0 does not yield 0^{\circ},180^{\circ} or 360^{\circ} as solution, that would be \sin x = 0. \cos x = 0 when x = 90^{\circ} and I'll let you find the other solution.
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    (Original post by thefatone)
    Same interval for x as last time
    0°≤x≤360°

    4cosx(cosx-1)=-5cosx

    i did
    4cos²x-4cosx=-5cosx
    4cos²x+cosx=0
    cosx(4cosx+1)=0

    but there's no answer in the back of the book which says that 0°,180°or 360° are answers
    You have done it correctly but cosx is not 0 when x=0,180 or 360
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    Thanks a bunch all i'll probably be posting loads more threads since i'll probably need more help on these xD
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    (Original post by thefatone)
    Thanks a bunch all i'll probably be posting loads more threads since i'll probably need more help on these xD
    I am too late to help?
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    (Original post by thefatone)
    Thanks a bunch all i'll probably be posting loads more threads since i'll probably need more help on these xD
    De nada.

    (Original post by TeeEm)
    I am too late to help?
    You snooze you lose
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    (Original post by Zacken)
    De nada.



    You snooze you lose
    I am glad you are doing more than your fair bit as I feel very "cannot be bothered" recently
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    My girl bled a bit. She had the best time tho in missionary, doggy style was orite too
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    (Original post by TeeEm)
    I am glad you are doing more than your fair bit as I feel very "cannot be bothered" recently
    No problemo! You deserve a break.
 
 
 
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