Question

if (sec x-2)(2sec x-1), then x terminates in which quadrant ?

Answer 1

2sec^2 x - 5sec x + 2
Use substitution.
sec x = y
2y^2 - 5y + 2 = 0
(2y - 1)(y - 2) = 0
Substitute back in.
2sec x - 1 = 0
sec x = 1/2

sec x - 2 = 0
sec x = 2
I'm pretty sure that it terminates in quadrant 1. If not, it's quadrant 4.

Answer 2

thank you