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    1st question:

    (1-cos2θ)/sin2θ ≡ tan θ
    θ≠(npi)/2

    solve, giving exact answers in terms of pi:
    2(1-cos2θ) = tanθ
    0<θ<pi


    2nd question:

    curve with equation y = ln 3x crosses the x-axis at the point P(1/3,0)
    the normal to the curve at the point Q, with x-coordinate q, passes through the origin.
    show that x=q is a solution of the equation x^2 + ln 3x = 0


    Thanks for any replies.

    Laura xxxx
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    Answers to both questions attached.
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    Hi - thanks for that.
    Sorry about the weird smilies - I forgot to disable them.

    Laura xxxx
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    you lost me on part b)
    could you try explaining it in a bit more detail for my "simple little brain - (but still doing A2 maths)" please?

    Thanks

    Laura xxxx
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    (Original post by Laura_M)
    you lost me on part b)
    could you try explaining it in a bit more detail for my "simple little brain - (but still doing A2 maths)" please?

    Thanks

    Laura xxxx
    Sorry about that.
    The first line is just a standard quote of the equation
    The second line is just a statement of the identity that we earlier proved, but in the form 1-cos2θ=...
    In the third line we substitute our expression for 1-cos2θ into the equation we wish to solve.
    In the fourth line we factorise out tanθ (notice we don't divide, as that would lose a solution) and the procedure is simple from there.

    I've actually made a mistake i think. If the inequality is 0<θ<pi then the 0 shouldn't be there. If the inequality is 0<=θ<=pi then pi should also be there. Mine only applies to 0<=θ<pi.
 
 
 
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