# math

posted by .

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

which is 2 pi - 1 and 2 pi - 1.18

also pi -1 and pi - 1.18

## Similar Questions

1. ### Math

Solve the equation for x in the interval 0<x<2pi 1/ 1+tan^2x = -cos x How would i do this?
2. ### PreCalculus

Hi I need some assistance on this problem find the exact value do not use a calculator cot[(-5pi)/12] Here is my attempt RT = square root pi = 3.14... cot[(-5pi)/12]={tan[(-5pi)/12]}^-1={tan[(2pi)/12-(8pi)/12]}^-1={tan[pi/6-(2pi)/3]}^-1={(tan[pi/6]-tan[(2pi)/3])/(1+tan[pi/6]tan[(2pi)/3])}^-1={((RT[3])/3+RT[3])/(1+((RT[3])/3)(-RT3))}^-1 …
3. ### Trig

Find all solutions of the equation on the interval [0,2pi): Tan^2x=1-secx

5. ### Trig

can I please have help with these 3 questions?
6. ### Math

Can I please get some help on these questions: 1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]?
7. ### trig

solve the equation on the interval [0, 2pi]: tan^2 x - 3 tan x + 2 = 0
8. ### Math (Pre-Calc)

Find all the solutions of the equation in the interval {o,2pi) 5sqrt3 tan x+3=8sqrt3 tan x
9. ### Trigonometric

tan^2 x − 7 tan x − 8 = 0 Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π).
10. ### math

Find all solutions to the equation tan(t)=1/tan (t) in the interval 0<t<2pi. Solve the equation in the interval [0, 2pi]. The answer must be a multiple of pi 2sin(t)cos(t) + sin(t) -2cos(t)-1=0 Find all solutions of the equation …

More Similar Questions