How many solutions does the equation $\displaystyle sin 2002x=sin 2003x$ have in the interval$\displaystyle [0,2\pi]$?

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- Nov 6th 2007, 12:21 AMperashHow many solutions
How many solutions does the equation $\displaystyle sin 2002x=sin 2003x$ have in the interval$\displaystyle [0,2\pi]$?

- Nov 6th 2007, 07:43 AMThePerfectHacker
$\displaystyle \sin 2002x = \sin 2003x \implies \sin 2003 x - \sin 2002 x = 0 \implies \cos \frac{4005x}{2} \sin \frac{x}{2} = 0$.

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

(2) If $\displaystyle \cos \frac{4005x}{2} = 0$ then $\displaystyle \frac{4005x}{2} = \frac{\pi}{2} + \pi n \implies 4005x = \pi + 2\pi n = \pi (2n+1)$.

So we want, $\displaystyle 0 \leq \frac{\pi (2n+1)}{4005} \leq 2\pi \implies 0\leq \frac{2n+1}{4005} \leq 2 $. Thus, $\displaystyle 0\leq 2n+1 \leq 8010$.

You can solve from here.