Thursday

November 20, 2014

November 20, 2014

Posted by **applebottom** on Friday, November 5, 2010 at 8:05pm.

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

- math -
**Damon**, Friday, November 5, 2010 at 8:43pmlet 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**

**Related Questions**

Math - Solve the equation for x in the interval 0<x<2pi 1/ 1+tan^2x = -...

trig - please help me with some questions I skipped on a review for our test ...

trig - solve the equation on the interval [0, 2pi]: tan^2 x - 3 tan x + 2 = 0

PreCalculus - Hi I need some assistance on this problem find the exact value do ...

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

Trig - can I please have help with these 3 questions? 1. Solve this equation ...

Math - Can I please get some help on these questions: 1. How many solutions does...

trig - solve the equation 2 tan C-3=3 tan X-4 algebraically for all values of C ...

calculus - I need help on finding the local linear approximation of tan 62 ...

Help me trig. - solve the equation in the interval [0,2pi) tan (t/2)-sint=0