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2005 Paper 2 Q8 - HSN forum

# 2005 Paper 2 Q8

### #1will_789

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Posted 16 May 2006 - 11:20 AM

2005 Paper 2 question 8
There a diagram, and you need to find out the point of intersection between ksin2x and sinx
Could someone please show me how to do this?

### #2Nathan

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Posted 16 May 2006 - 11:45 AM

you can say that the curves intersect where y=y

hence;

ksin2x=sinx
ksin2x-sinx=0
2ksinxcosx-sinx=0
sinx(2kcosx-1)=0

sinx=0 or..2kcosx-1=0

where sinx=0, x=0, or 2[pi] O(0,0) , B(,0) and D(2 ,0)

where;
2kcosx-1 = 0;
cosx = 1/2k

since we know what B and D are, it can be implied that cosx = 1/2k at the points A and C

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