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- mathsyperson
- 2005-12-06 02:13:11
Correct, but you've missed out some possible solutions.
π/12, 5π/12, 3π/4, 13π/12, 17π/12 and 7π/4 all work.
- krassi_holmz
- 2005-12-05 17:20:46
2sin3xcos3x=2(sin6x)/2=sin6x, so
sin6x=1
x = (1/6)(Pi/2)=Pi/12
- runnerboy16
- 2005-12-05 15:57:28
thanks, but that isnt what i need.
- Ricky
- 2005-12-05 15:47:25
http://aleph0.clarku.edu/~djoyce/java/trig/identities.html
Use the product identity, about 3/4th the way down the page.
- runnerboy16
- 2005-12-05 15:32:33
I need to know how to find all the solutions for the problem- 2sin3xcos3x=1 within the restrictions of [0,2pie)