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You are not logged in. #1 20051205 15:32:33
Need Help FastI need to know how to find all the solutions for the problem 2sin3xcos3x=1 within the restrictions of [0,2pie) #2 20051205 15:47:25
Re: Need Help Fasthttp://aleph0.clarku.edu/~djoyce/java/trig/identities.html "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 20051205 17:20:46
Re: Need Help Fast2sin3xcos3x=2(sin6x)/2=sin6x, so Last edited by krassi_holmz (20051205 17:21:31) IPBLE: Increasing Performance By Lowering Expectations. #5 20051206 02:13:11
Re: Need Help FastCorrect, but you've missed out some possible solutions. Why did the vector cross the road? It wanted to be normal. 