Question

Solve 4cos^2A = 3cosA for 90º≤A≤180º. (Enter only the number.)





Find A when 0º≤A≤90º and 3tan^2A = 2tanA + 1. (Enter only the number.)

Answer 1

let z = cos A

4 z^2 - 3 z = 0
z(4z-3) = 0
z = 0 or z = 3/4
so
cos A = 0 when z = 90 degrees
cos A = 3/4 when z = 41.4 degrees but that is not between 90 and 180

Answer 2

4cos^2 A - 3cosA = 0
cosA(4cosA - 3) = 0
cosA=0 or cosA = 3/4

but your domain is quadrant II, where the cosine is negative
except cos90° = 0
so x = 90°

do the 2nd the same way by factoring out tanA