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Trig identies

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data
    2cos2x-2cosx...how do you simplify this further?


    2. Relevant equations



    3. The attempt at a solution
    2(cos2x-cosx)..but i have to find 0=2cos2x-2cosx so this doesnt really help me.
     
  2. jcsd
  3. Jul 21, 2009 #2

    Office_Shredder

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    You're trying to find when 2cos(2x)-2cos(x)=0? Start by what you did... you want to find x such that cos(2x)=cos(x). Hint: cosine is periodic
     
  4. Jul 21, 2009 #3
    would it be pi/2 and 3pi/2?
     
  5. Jul 21, 2009 #4
    nope....maybe try drawing cos(2x) and cos(x)?
     
  6. Jul 21, 2009 #5

    zcd

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    [tex]\cos(2x)=2\cos^{2}(x)-1[/tex], so [tex]2\cos(2x)-2\cos(x)=2\cos^{2}(x)-1\cos(x)-1=0[/tex]. Factoring yields [tex](2\cos(x)+1)(\cos(x)-1)=0[/tex]
     
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