Finding the Zeroes of a Secant Graph

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Homework Help Overview

The discussion revolves around finding the zeroes of the function y = 2 sec(-2x + 180°) + 3. Participants are exploring the algebraic methods to determine the x-values where the function equals zero.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the equation to isolate the secant function and find its inverse. There is a discussion on the implications of using the inverse secant function and the conditions under which it is defined.

Discussion Status

The conversation is ongoing, with participants sharing their attempts and questioning the validity of certain steps in their reasoning. Some have provided insights into the nature of the secant function and its inverses, while others express confusion about the algebraic manipulation involved.

Contextual Notes

There is a noted concern regarding the behavior of the secant function and its inverse, particularly in relation to the values being considered. Participants are also reflecting on the limitations of algebraic methods compared to graphical approaches.

Sabellic
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This is a continuation of my previous post, so please bear with me

Homework Statement


Find the zeroes of the following function:

y= 2 sec (-2x+180deg) + 3

Homework Equations



Break down the equation into:

y= 2 sec -2(x-90deg) + 3;

Finding zeroes, means finding the x values when y=0, therefore:

0=2 sec [-2(x-90deg)]+3

The Attempt at a Solution



0=2 sec [-2(x-90deg)]+3

-3 = 2 sec [-2(x-90deg)]

-1.5 = sec [-2(x-90deg)]

-1.5/sec = [-2(x-90deg)]

^ this is where I get stuck. Because the inverse of sec or cos (sec^-1; cos^-1) is Error or Undefined. So the whole thing explodes. Now, I can tell when the graph crosses the x-axis (and therefore has a y=0 value) on a Graphing tool, but how can I find this out algebraically?

Thanks for the assistance.
 
Last edited:
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Sabellic said:
This is a continuation of my previous post, so please bear with me

Homework Statement


Find the zeroes of the following function:

y= 2 sec (-2x+180deg) + 3


Homework Equations



Break down the equation into:

y= 2 sec -2(x-90deg) + 3;

Finding zeroes, means finding the x values when y=0, therefore:

0=2 sec [-2(x-90deg)]+3



The Attempt at a Solution



0=2 sec [-2(x-90deg)]+3

-3 = 2 sec [-2(x-90deg)]

-1.5 = sec [-2(x-90deg)]

-1.5/sec = [-2(x-90deg)]
Now that's just silly. "sec" is NOT a number, sec[-29x- 90)] is NOT just "sec" times [-29x- 90]. You cannot just divide by "sec". You get rid of the secant function by using its inverse, arcsec(x). -2(x- 90)= arcsec(-1.5). Since secant of 1/cosine and cosine is always between - 1 and 1, secant= 1/cosine is between -1 and -infinity or 1 and infinity. arcsec(x)= sec-1(x) is definitely defined for x= -1.5.

^ this is where I get stuck. Because the inverse of sec or cos (sec^-1; cos^-1) is Error or Undefined. So the whole thing explodes. Now, I can tell when the graph crosses the x-axis (and therefore has a y=0 value) on a Graphing tool, but how can I find this out algebraically?

Thanks for the assistance.
 
editted
 
Last edited:
This is what I am getting:
-2(x- 90)= arcsec(-1.5)

-2(x- 90)=0.999657325
x-90=-0.499828662
x=89.5
 

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