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    5. Find, giving your answer to 3 significant figures where appropriate, the value of x for which:

    (c) ln sin x = –ln sec x, in the interval 0 < x < 90°.

    [ANSWER FROM MARK SCHEME:
    sec x = 1/ cos x
    sin x = cos x ⇒ tan x = 1 x = 45]

    >>>just wondering how to get to the second line please?
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    If you *e* both sides, it gets rid of the Ln's


    say you had lnx = 2, that would equal x = e^2
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    Ln sinx + Ln secx = 0 (bringing Ln secx to other side)

    Law of logirithms addition thing

    Ln (sinx * 1/cosx) = 0

    Take exponent from both side

    Sinx/Cosx = 1
    Tanx = 1
    x = 45

    Please if you have the markscheme in electronic form, may you upload it somewhere or send it to me please :X =)
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    thanks both for the help..much appreciated!

    (Original post by LT0918)
    Please if you have the markscheme in electronic form, may you upload it somewhere or send it to me please :X =)
    yeh i have the mark scheme..perhaps pm me ur email n i'l send it over!?
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    Received - thanks very much

    For others the paper can be found here:

    http://www.apxc57.dsl.pipex.com/04%2...k%20scheme.pdf
 
 
 
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