# Finding the Zeroes of a Secant Graph

1. Oct 13, 2008

### Sabellic

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

1. The problem statement, all variables and given/known data
Find the zeroes of the following function:

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

2. Relevant 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

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: Oct 13, 2008
2. Oct 13, 2008

### HallsofIvy

Staff Emeritus
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.

3. Oct 13, 2008

### Sabellic

editted

Last edited: Oct 14, 2008
4. Oct 13, 2008

### Sabellic

This is what I am getting:
-2(x- 90)= arcsec(-1.5)

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