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January 19th, 2013, 03:03 PM  #1 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Confusing PreCalculus Problem. Can Anyone Counter This?
Question: Sin 4x = 2 Sin 2x What I did: = 2 (sin2x)(cos2x) = 2(2sinx cosx) Double angle formula = 2(2sinx cosx) (12sin^2x) =2(2 sinx cosx) Double angle formula =2sinx cosx (1 2sin^2x) = 2(sinx cosx) Divided everything by 2 = (12sin^2x) = 1 Divided both sides by 2 sinx cosx =22sin^2 x = 0 Added 1 on both sides. =2(1sinx)(1+sinx) = 0 Factored 1 sinx = 0 & sinx + 1 =0 Obvious method sinx = 1 & sinx = 1 " x= pi/2 U x = 3pi/2 Unit circle. But however, the answer key says this is wrong. Others' solutions: Sin 4x = 2 sin 2x = 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula = 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x 2sin 2x = 0 & cos2x = 1 sin 2x = 0 & cos 2x = 1 0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2 x= 0, pi, pi/2, 3pi/2 Does anyone know what I did wrong? ( the first solution) 
January 19th, 2013, 03:23 PM  #2  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Confusing PreCalculus Problem. Can Anyone Counter This? Quote:
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January 19th, 2013, 05:33 PM  #3  
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Re: Confusing PreCalculus Problem. Can Anyone Counter This? Quote:
Quote:
Oh, okay. Thanks!  
January 19th, 2013, 05:48 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Confusing PreCalculus Problem. Can Anyone Counter This?
I would use the strategy suggested by [color=#0000FF]HallsofIvy[/color]: Use the doubleangle identity for sine on the left, and arrange as: Factor: We have two cases to consider (where ): i) ii) However, we see that all of the solutions from the second case are part of the first, so we may simply state: 

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2cos(2x)^2  cos2x  1 = 0 why can't i use a double angle identity,sinx cosx divided by sinxcosx=1 2sinxcosx divided by 2sin^2x1,formula of 2 sinx devited by2,mathematics 2sinx formula
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