Is the Trigonometry Book Wrong? Conflicting Answers for x in secx^2-4=0

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Liger20
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b]1. Homework Statement [/b]
Hey, I have a problem that I worked out in a trigonometry book. I got different answers from what the book gave, but I think that I am right and the book is wrong. Could someone please tell me who is right?


Homework Equations



Okay, they gave me the following problem: solve for x in secx^2-4=0 when 0 is less than or equal to x, which is less than or equal to 360.

Here is how the book solves it:
(secx-2)(secx+2)

secx-2=0 secx+2=0
secx=2 secx=-2
x=sec^-1(2) x=sec^-1(-2)
= 30, 330 = 150, 210

Okay, here's where I have a big problem: secant is the reciprocal function of cosine, right?
So if we evaluate x=sec^-1(2), we get cos^-1(1/2), which does not equal 30 and 330 when plugged into the calculator. Same situation with x=sec^-1(-2).
I am pretty sure that this is a horrible typo, but I would be delighted to be corrected otherwise.





The Attempt at a Solution


 
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I agree with you. If you want places where cos(x) = +/-.5 then that is 60,120,240,300. Also I graphed [sec(x)]^2-4 and it was 0 at those places. The values they gave were correct for sine. So unless the original problem is [csc(x)]^2-4=0, their answer is wrong.
 
Actually sec x = 1/cos x. I think you are confusing ^-1 and inverse functions.
 
I think that there is a typo, the answers should actually be x = 60, 120, 240, 300.

Solution from where the book leaves off:
sec(x) = 2, sec(x) = -2
1/2 = cos(x), -1/2 = cos(x)
Use graph or calculator to find the angles are those that I stated above.
 
dlgoff said:
Actually sec x = 1/cos x. I think you are confusing ^-1 and inverse functions.
No, he said "reciprocal", not "inverse". There is NO cos-1(2).