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Math Help - How many solutions

  1. #1
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    How many solutions

    How many solutions does the equation sin 2002x=sin 2003x have in the interval  [0,2\pi]?
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  2. #2
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    Quote Originally Posted by perash View Post
    How many solutions does the equation sin 2002x=sin 2003x have in the interval  [0,2\pi]?
    \sin 2002x = \sin 2003x \implies \sin 2003 x - \sin 2002 x = 0 \implies \cos \frac{4005x}{2} \sin \frac{x}{2} = 0.

    (1) If \sin \frac{x}{2} = 0 then \frac{x}{2} = \pi n \implies x = 2\pi n since we require that x\in [0,2\pi] it means x=0,2\pi.

    (2) If \cos \frac{4005x}{2} = 0 then \frac{4005x}{2} = \frac{\pi}{2} + \pi n \implies 4005x = \pi + 2\pi n = \pi (2n+1).
    So we want, 0 \leq \frac{\pi (2n+1)}{4005} \leq 2\pi \implies 0\leq \frac{2n+1}{4005} \leq 2 . Thus, 0\leq 2n+1 \leq 8010.
    You can solve from here.
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