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Thread: Solving for zeros of tan(x) function with transformation

  1. #1
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    Solving for zeros of tan(x) function with transformation

    I have the equation y = -2 tan (3x + 180) + 3

    I need to find the x-intercepts. To do this, I know that I need to set y to equal 0 and then solve for x.

    This is what I've done:

    y = -2 tan (3x + 180) + 3

    -3 = -2 tan (3x + 180)

    3/2 = tan (3x + 180)

    At this point I'm stuck. I don't know how to further isolate the x. Can I do this??

    tan ^-1 3/2 = 3x + 180, solve for tan ^-1 3/2 and continue isolating x?

    Thanks!!
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  2. #2
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    Assuming that you meant $\displaystyle \displaystyle \tan^{-1}{\left(\frac{3}{2}\right)} = 3x + 180$, then yes, you have the right idea.

    However, $\displaystyle \displaystyle \tan^{-1}{\left(\frac{3}{2}\right)}$ does not have an exact surd value, so if you are asked to keep your answer exact, you would not give a decimal approximation for this.
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    Okay, great... but how do I list zeros for this then?

    Or, do I just write:

    Roots: x = (tan^-1 (3/2) - 180)/3 ?

    When I solve for x using the decimal value of tan^-1 (3/2) I get -41.23002251, but that is only one root, and there has to be more than one.

    Can I take that number and just add intervals of 60 degrees since the vertical asymptotes are set at intervals of 60 degrees?
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    You first need to recall that the tangent function is periodic, with a period of $\displaystyle \displaystyle 180^{\circ}$. So

    $\displaystyle \displaystyle \begin{align*}\tan{(3x+180)} &= \frac{3}{2} \\ 3x + 180 &= \tan^{-1}{\left(\frac{3}{2}\right)} + 180n \textrm{ where }n \in \mathbf{Z} \\ 3x &= \tan^{-1}{\left(\frac{3}{2}\right)} + 180n - 180 \\ x &= \frac{1}{3}\tan^{-1}{\left(\frac{3}{2}\right)} + 60n - 60\end{align*}$
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    Okay, but how do I solve that for the x-intercepts?
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    Those ARE the x-intercepts.
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    Lol, I think I'm still lost.... so I just list x = tan^-1 (3/2) + 60n - 60 for the x-intercepts?? Does that account for all of them then? I can't use that equation to solve for and list specific intercepts?
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  8. #8
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    Re: Solving for zeros of tan(x) function with transformation

    Yes; all the x-intercepts would be given by $\displaystyle x = \tan^{-1}(3/2) + 60n - 60$, because remember, $\displaystyle n $ is any integer. So you would get a different solution for each different value of $\displaystyle n$.

    To get specific x-intercepts, you would just need to specify what $\displaystyle n$ you are using. For example, $\displaystyle n = 2$ gives:

    $\displaystyle x = \tan^{-1}(3/2) + 60(2) - 60 = tan^{-1}(3/2) + 60$.

    I hope that this helps.
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