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how do i find the exact value of cos x when 8cos x cosec^2 x = 3 watch

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    how do i find the exact value of cos x when 8cos x cosec^2 x = 3
    i know that its 1/3 but not sure in teh slightest on how to get there. this was in the last C3 paper i did
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    from what i remember( wierd enough) cosec^2x = 1-sec^2x or something similar there is a formula... and the sec^2x has a formula probaly which you change to cosx then you bring the 3 on to the left and factorise...
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    (Original post by ada1023)
    how do i find the exact value of cos x when 8cos x cosec^2 x = 3
    i know that its 1/3 but not sure in teh slightest on how to get there. this was in the last C3 paper i did
    from what i remember( wierd enough) cosec^2x = 1-sec^2x or something similar there is a formula... and the sec^2x has a formula probaly which you change to cosx then you bring the 3 on to the left and factorise...
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    this was the ocr exam question today wasnt it
    i got stuck on it for aaaages but i think (hope it worked out ok)

    cosec^2 = 1+ cot^2
    sub that in
    8cosx X 8(cos^2/sin^2)=3
    8cosx X 8(cos^2/1-cos^2)=3

    8cosx-8cos^3 + 8cos^3 = 3 - 3cos^2

    3cos^2+8cos-3=0
    quadraticify it
    (3cos-1)(cos+3)=0
    cos = 1/3 as cos =-3 is undefined

    hhope this is right and it helps (obviously cant do much about it now good luck)
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    I got this right as well hehe, this paper is well hard I [rainbow]******[/rainbow] up so many questions
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    thanks alot, but between these 2 steps im a bit confused, can you break it down a bit more? (sorry im a bit slow :P)

    8cosx X 8(cos^2/1-cos^2)=3

    8cosx-8cos^3 + 8cos^3
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    ah this is a clever question;

    8cosxcosec^2x=3

    \frac{8cosx}{sin^2x}=3

    8cosx=3sin^2x

    8cosx=3(1-cos^2x)

    8cosx=3-3cos^2x

    3cos^2x+8cosx-3=0

    (3cosx-1)(cosx+3)=0

    cosx=-3, \frac{1}{3}

    -1<cosx<1

    so, cosx=\frac{1}{3}
 
 
 
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