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Thread: Find x-intercepts

  1. #1
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    Find x-intercepts

    Hi
    I am having trouble finding the x-intercepts from $\displaystyle -2\pi to 2\pi$
    from the equation: $\displaystyle y=2sin(x-\frac{\pi}{4})+1$

    This is what i have done but it doesn't match the book's answers:

    $\displaystyle 0=2sin(x-\frac{\pi}{4})+1$
    $\displaystyle \frac{-1}{2}=sin(x-\frac{\pi}{4})$
    $\displaystyle x-\frac{\pi}{4}=\pi+\frac{\pi}{6},2\pi-\frac{\pi}{6},-\pi+\frac{\pi}{6},-2\pi-\frac{\pi}{6}$
    $\displaystyle x=\frac{7\pi}{6}+\frac{\pi}{4},\frac{11\pi}{6}+\fr ac{\pi}{4},\frac{-5\pi}{6}+\frac{\pi}{4},\frac{-13\pi}{6}+\frac{\pi}{4}$
    $\displaystyle \frac{17\pi}{12},\frac{25\pi}{12},\frac{-7\pi}{12},\frac{17\pi}{8}$

    P.S
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  2. #2
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    Whilst your answers are technically correct, 2 of them do not satisfy the limits. This is because your limits were wrong when solving $\displaystyle \sin(x - \frac{\pi}{4}) = \frac{-1}{2} $

    Be careful when you are solving questions like this. If the angle is $\displaystyle \theta + k $, the limits you need to apply when solving for $\displaystyle \theta + k $ changes. If you had limits $\displaystyle -2\pi \le \theta \le 2\pi $, the limits for $\displaystyle \theta + k $ should be $\displaystyle -2\pi + k \le \theta + k \le 2\pi + k $
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  3. #3
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    So it should be like this:
    $\displaystyle \pi+\frac{\pi}{6},-\pi+\frac{\pi}{6},2\pi+\frac{\pi}{6},-2\pi+\frac{\pi}{6}$

    However for the last two i get the incorrect answers.
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  4. #4
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    How did you figure the last two solutions?

    General solution for $\displaystyle x - \frac{\pi}{4} $ is:

    $\displaystyle x - \frac{\pi}{4} = n\pi + (-1)^n \left(\frac{-\pi}{6}\right) $

    Limits: $\displaystyle -2\pi - \frac{\pi}{4} \le x - \frac{\pi}{4} \le 2\pi - \frac{\pi}{4} $

    For $\displaystyle n = 0 \Rightarrow \left(\frac{-\pi}{6}\right) $
    For $\displaystyle n = 1 \Rightarrow \pi - \left(\frac{-\pi}{6}\right) = \pi + \left(\frac{\pi}{6}\right) $
    For $\displaystyle n = -1 \Rightarrow -\pi - \left(\frac{-\pi}{6}\right) = -\pi + \left(\frac{\pi}{6}\right)$
    For $\displaystyle n = 2 \Rightarrow 2\pi + \left(\frac{-\pi}{6}\right) = 2\pi - \left(\frac{\pi}{6}\right) $ but this exceeds the limit, so it's discounted.
    For $\displaystyle n = -2 \Rightarrow -2\pi + \left(\frac{-\pi}{6}\right) = -2\pi -\left(\frac{\pi}{6}\right) $


    Hence $\displaystyle x = \frac{\pi}{12} ; \frac{17\pi}{12} ; \frac{-7\pi}{12} ; \frac{-23\pi}{12}$
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  5. #5
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    Quote Originally Posted by Gusbob View Post
    How did you figure the last two solutions?
    yeh, i made the denominator 24 and not 12.
    I tried it again with 12 and it was correct.
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