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#1 2005-12-05 15:32:33

runnerboy16
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Need Help Fast

I need to know how to find all the solutions for the problem- 2sin3xcos3x=1   within the restrictions of [0,2pie)

#2 2005-12-05 15:47:25

Ricky
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Re: Need Help Fast

http://aleph0.clarku.edu/~djoyce/java/trig/identities.html

Use the product identity, about 3/4th the way down the page.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#3 2005-12-05 15:57:28

runnerboy16
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Re: Need Help Fast

thanks, but that isnt what i need.

#4 2005-12-05 17:20:46

krassi_holmz
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Re: Need Help Fast

2sin3xcos3x=2(sin6x)/2=sin6x, so
sin6x=1
x = (1/6)(Pi/2)=Pi/12

Last edited by krassi_holmz (2005-12-05 17:21:31)


IPBLE:  Increasing Performance By Lowering Expectations.

#5 2005-12-06 02:13:11

mathsyperson
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Re: Need Help Fast

Correct, but you've missed out some possible solutions.

π/12, 5π/12, 3π/4, 13π/12, 17π/12 and 7π/4 all work.


Why did the vector cross the road?
It wanted to be normal.

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