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

1. ## 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!!

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

However, $\displaystyle \tan^{-1}{\left(\frac{3}{2}\right)}$ 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.

3. 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?

4. You first need to recall that the tangent function is periodic, with a period of $\displaystyle 180^{\circ}$. So

\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*}

5. Okay, but how do I solve that for the x-intercepts?

6. Those ARE the x-intercepts.

7. 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?

8. ## Re: Solving for zeros of tan(x) function with transformation

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

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

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

I hope that this helps.