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# Can someone help me out, please ? watch

1. 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
2. Answers to both questions attached.
Attached Images

3. Hi - thanks for that.
Sorry about the weird smilies - I forgot to disable them.

Laura xxxx
4. 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
5. (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
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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