Finding Points of discontinuity

[mikagurl]

Hi all. My problem is to find the points of discontinuity of the function 1/2-4cos(3z). I do not know how to solve this but I started by subtracting the 2 from both sides. I now have 4cos(3z)=-2. I then divided both sides of the equation -4. Now I have cos(3z)=1/2. I think I should take cos inverse of (1/2) but I am not sure. Also what should I do with the 3z at this point.

 

[pka]

Hi all. My problem is to find the points of discontinuity of the function 1/2-4cos(3z).

There is something wrong with the statement of the question.
If the function is $\displaystyle f(z)=\frac{1}{2}-4\cos(3z)$ then it is continuous everywhere.

 

[stapel]

My problem is to find the points of discontinuity of the function 1/2-4cos(3z).

As posted, the function (in terms of the variable "z"?) is as follows:

. . . . .$\displaystyle f(z)\, =\, \dfrac{1}{2}\, -\, 4\cos(3z)$

...which obviously wouldn't have any issues with discontinuity. Do you perhaps mean the following?

. . . . .$\displaystyle f(z)\, =\,\dfrac{1}{2\, -\, 4\cos(3z)}$

I started by subtracting the 2 from both sides. I now have 4cos(3z)=-2.

You would only have this result if you'd started with f(z) = 0. But you didn't; you're not solving for the zeroes of the function. Instead, you started with f(z) = 1/2 - 4cos(3z) -- or perhaps f(z) = 1/[2 - 4cos(3z)]. By subtracting 2 from "both sides", you now have f(z) - 2 = -3/2 - 4cos(3z). What does this accomplish?

Kindly please reply with clarification of what the original function is. When you reply, please explain what you are attempting to do. For instance, did I guess correctly what the function was meant to be? If so, are you "setting the denominator equal to zero" and then attempting to find the zeroes of the denominator in an effort to find the points where the function is not defined?

Thank you!

 

[mikagurl]

The function is 1/(2-4cos(3z)). I have to find the points of discontinuity. Yes, I did set the equations = to zero. Sorry for the confusion.

 

[stapel]

The function is 1/(2-4cos(3z)). I have to find the points of discontinuity. Yes, I did set the equations = to zero.

I will assume that, by "the equations [plural] = to zero", you mean "the denominator [singular] equal to zero". If so, then your methodology is correct. You then have 2 - 4cos(3z) = 0, so 2 = 4cos(3z), which simplifies as 1/2 = cos(3z).

Now think back to what you learned back in trigonometry (or "pre-calculus", if that's what they called the course you took before calc). For what angle value(s) of @ ("theta") is cos(@) equal to one-half? Then for what angle values of 3z is cos(3z) equal to one-half? Then what are the solutions values for z?