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#1 2005-12-05 15:32:33
- runnerboy16
- Novice
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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
- Moderator
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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..."
#4 2005-12-05 17:20:46
- krassi_holmz
- Real Member
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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.