Wednesday
March 29, 2017

Post a New Question

Posted by on .

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

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

  • math - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question