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