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**runnerboy16****Member**- Registered: 2005-12-04
- Posts: 3

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

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

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**runnerboy16****Member**- Registered: 2005-12-04
- Posts: 3

thanks, but that isnt what i need.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

2sin3xcos3x=2(sin6x)/2=sin6x, so

sin6x=1

x = (1/6)(Pi/2)=Pi/12

*Last edited by krassi_holmz (2005-12-04 18:21:31)*

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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