Solve the equation in the interval [0,2pi]. List all solutions

tan^2(x)+0.8 tan(x)-3.84 = 0

1 answer

let z = tan x

z^2 +.8 z -3.84 = 0

z = [ -.8 +/- sqrt (.64 +15.36)]/2

z = [ -.8 +/- sqrt (16) ] / 2

z = [ -.8 +/- 4 ]/2

z = tan x = -2.4 or -1.6

tan is negative in quadrants 2 and 4
so

about -1 radian or about -1.18 radian
which is 2 pi - 1 and 2 pi - 1.18

also pi -1 and pi - 1.18