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September 13th, 2009, 04:24 PM  #1 
Member Joined: Feb 2008 Posts: 30 Thanks: 0  Solve X where infinity>x>infinity
Sinx + 1 = 2Cos^2x I got pi/6 + 2piK 5pi / 6 + 2piK pi/2 + 2pik but it says pi/2 + 2pik isnt an answer? WHY? 
September 13th, 2009, 05:05 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 723  Re: Solve X where infinity>x>infinity Quote:
 
September 13th, 2009, 05:15 PM  #3 
Member Joined: Feb 2008 Posts: 30 Thanks: 0  Re: Solve X where infinity>x>infinity
Using trig identities i made it Sinx + 1 = 2  2Sin^2x Sin^2x + Sinx  1 = 0 (2Sinx1)(Sinx+1)=0 Sinx= 1/2, 1 
September 13th, 2009, 06:54 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
Your second line above should have been 2Sin^2x + Sinx  1 = 0, but everything else you did is correct. Assuming the problem was given correctly, your textbook (or whatever) seems to be incorrect; also, mathman made a slip. 
September 13th, 2009, 07:14 PM  #5 
Member Joined: Feb 2008 Posts: 30 Thanks: 0  Re: Solve X where infinity>x>infinity
Yes, it is 2 sin^2x.. typo. For some reason I have a feeling theres a reason why pi/2 + 2piK wasn't included.

September 14th, 2009, 05:13 AM  #6 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 
If the original problem were then would not be a solution (I hope you see why). Is this relevant, or not?


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infinity>x>infinity, solve 
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