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Trig transformation Question

  1. Apr 8, 2007 #1
    The question is "Solve each equation for exact solutions over the interval of [0,2pi).

    Equation: sin x = sin 2x

    I transformed the equation to

    sin x = 2sin x cos x

    then

    sin x/2sin x = cos x

    cos x = 1/2

    Answer: pi/3, 5pi/3

    but the book also has the answers 0, pi.

    What must I do to find the other two solutions?
     
  2. jcsd
  3. Apr 8, 2007 #2

    arildno

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    Dearly Missed

    You ASSUMED that sin(x) was non-zero when you divided with it, since you can't divide by zero.

    Since we see that the equation is ALSO fulfilled in the case sin(x)=0
    (The equation reduces then to 0=2*0*cos(x) which is always true),

    then the solutions of sin(x)=0 are also solutions of the original equation.
    Guess what those numbers might be..:smile:
     
  4. Apr 8, 2007 #3
    Thanks for the explanation. :)

    *edit* Figured it out now.
     
    Last edited: Apr 8, 2007
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