Posted by applebottom on .
Solve the equation in the interval [0,2pi]. List all solutions
tan^2(x)+0.8 tan(x)3.84 = 0

math 
Damon,
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