# Thread: Roots of sec(x) function with transformation

1. ## Roots of sec(x) function with transformation

I was able to solve a similar question finding the roots of the tan(x) function with a transformation, but this one seems to be giving me a bit more trouble...

The equation is y = 2sec(-2x + 180) + 3

I know to find the roots I have to set y to 0 and solve for x.

This is what I've done so far:

y = 2sec(-2x + 180) + 3
0 = 2sec (-2x + 180) + 3
-3 = 2sec (-2x + 180)
-3/2 = sec (-2x + 180)
sec^-1 (-3/2) = -2x + 180

This is where I'm running into trouble. I know that sec = 1/cos, so I rewrote this line as:

1/(cos^-1 (-3/2) ) = -2x + 180

But I'm not able to solve for cos^-1 (-3/2) so now I'm at a bit of a loss.

Any help would be greatly appreciated. Thanks so much!!!

2. Originally Posted by starshine84
I was able to solve a similar question finding the roots of the tan(x) function with a transformation, but this one seems to be giving me a bit more trouble...

The equation is y = 2sec(-2x + 180) + 3

I know to find the roots I have to set y to 0 and solve for x.

This is what I've done so far:

y = 2sec(-2x + 180) + 3
0 = 2sec (-2x + 180) + 3
-3 = 2sec (-2x + 180)
-3/2 = sec (-2x + 180)
Correct up to here.

sec^-1 (-3/2) = -2x + 180

This is where I'm running into trouble. I know that sec = 1/cos, so I rewrote this line as:

1/(cos^-1 (-3/2) ) = -2x + 180

But I'm not able to solve for cos^-1 (-3/2) so now I'm at a bit of a loss.
You need to take the reciprocal of both sides (you only did the right hand side)

$-\dfrac{3}{2} = \sec (-2x + 180) \rightarrow -\dfrac{2}{3} = \cos (-2x+180)$

3. Okay, so from there, is this right?

-2/3 = cos(-2x + 180)
cos^-1 (-2/3) = cos(-2x + 180)
-48.1896851 = -2x
24.09484255 = x

4. You can test potential solutions by seeing if they work in the original equation.

This one is a root but it's only the principle root - since $\sec(x)$ is periodic you're going to have infinite roots unless the question specifies a range $(0 \leq x \leq 360)$ is common

5. Okay, I think I'm on the right track now. Thanks for you help!!