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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
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Re: Solve X where -infinity>x>infinity

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 Originally Posted by -DQ- 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?
The equation as written doesn't seem right. Your first 2 solutions do not satisfy the equation, while the third one does.

 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 (2Sinx-1)(Sinx+1)=0 Sinx= 1/2, -1
 September 13th, 2009, 06:54 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 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 $\frac{\sin x+1}{2\cos^2x}=1,$ then $-\pi/2+2\pi k$ would not be a solution (I hope you see why). Is this relevant, or not?

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