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Solving equations

  1. Jan 8, 2008 #1
    Suppose you are solving equations in the interval 0<=xx<=2pi....Without actually solving equations, what is the difference between the number of solutions of sinx=1 and sin2x=1? How do you account for this difference
  2. jcsd
  3. Jan 8, 2008 #2

    Gib Z

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    Homework Helper

    Well since sin 2x = 2 sin x cos x = 1, and you also want solutions to sin x =1, we can see that the number of solutions of sin x = sin 2x = 1 is the number of times cos x is equal to 1 in the interval.
  4. Jan 8, 2008 #3
    [itex]sin x = 1[/itex]
    [itex]sin x = sin (\frac{\pi}{2})[/itex]
    [itex]x = n\pi + (-1)^n (\frac{\pi}{2})[/itex]

    But, as [itex]x \in [0, 2\pi][/itex]

    Hence, select the values for n ([itex]n \in N[/itex]), such that [itex]x \in [0, 2\pi][/itex].

    The satisfying values are: [itex]n \in \{0, 1\}[/itex] Put this values for x, and you shall get [itex]x \in \{\frac{\pi}{2}\}[/itex]. This is because when we have [itex]\alpha[/itex] as [itex]\frac{\pi}{2}[/itex], we get the same solutions for [itex]n = 0; n = 1[/itex].

    Do the same for 2x ([itex]x \in [0, 2\pi][/itex]), ([itex]2x \in [0, 4\pi][/itex]) and you shall get the solutions for 2x. It again gives us:

    [itex]2x = n\pi + (-1)^n (\frac{\pi}{2})[/itex]

    Here, {0, 1, 2, 3} satisfy 'n', giving 2 unique solutions i.e. [itex]x \in \{\frac{\pi}{4}, \frac{5\pi}{4}\}[/itex].
    Last edited: Jan 9, 2008
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