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
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?