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