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September 13th, 2009, 04:24 PM   #1
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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?
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September 13th, 2009, 05:05 PM   #2
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Re: Solve X where -infinity>x>infinity

Quote:
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.
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September 13th, 2009, 05:15 PM   #3
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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
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September 13th, 2009, 06:54 PM   #4
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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.
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September 13th, 2009, 07:14 PM   #5
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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.
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September 14th, 2009, 05:13 AM   #6
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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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