If $\displaystyle (sec x - 2)(2 sec x - 1) = 0 $ , then x terminates in:a) Quadrants I and II, only

b) Quadrants I and IV, only

c) Quadrants I, II, III, and IV

d) Quadrant I, only

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- Apr 29th 2009, 02:26 PMThaBerettaX Terminates in What Quadrant?
*If $\displaystyle (sec x - 2)(2 sec x - 1) = 0 $ , then x terminates in:*a) Quadrants I and II, only

b) Quadrants I and IV, only

c) Quadrants I, II, III, and IV

d) Quadrant I, only - Apr 29th 2009, 02:28 PMVonNemo19hello
Solve for x, and determine the domain my son.(Wink)

- Apr 29th 2009, 02:34 PMThaBeretta
Hmm...I'm really not sure how to go about solving for x. Also, how would I determine the domain once I have x?

- Apr 29th 2009, 02:41 PMVonNemo19
When you solve a quadratic equation, in some cases you factor. Right.

Well your problem looks a lot like something like this: (X+a)(X+b)=0

and the solutions are: x=-a, x=-b

Can you see my meaning? - Apr 30th 2009, 03:31 AMThaBeretta
- Apr 30th 2009, 03:43 AMSundae
Let secx = y

therfore, y = 2 & y = 1/2

Because both values are positive they are in the 1st and 4th quadrant.

in the first quadrant everything is positive, in second quadrant only sin/cosec is positive, in the third quadrant only tan/cot is positive and in 4th quadrant cos/sec is positive.

Is that what u were askin? - Apr 30th 2009, 09:25 AMThaBeretta
So the correct answer is choice "b"?

- Apr 30th 2009, 09:30 AMVonNemo19yes
yes

- Apr 30th 2009, 09:58 AMThaBeretta
Okay, thank you.