# Thread: Trig equation with tangents

1. ## Trig equation with tangents

Here's the problem: tan^2x-3tanx-4=0, x [0, 3pi]

What I've worked out so far:

(tanx+1)(tanx-4)=0
tan(x)=-1, tan(x)=4
arctan(-1)=x, arctan(4)=x
-0.7853982=x, 1.325818=x

I'm aware the period is 360 degrees... do I simply add 360 to both solutions and then check to see which ones are correct? How do I use the "tangent is positive in the first and fourth quadrants" to solve?

2. Be careful with radians versus degrees there. Which mode was your calculator in?

3. Radians - I double checked

4. The period for the tangent is 180 degrees!
And the tangent is positive in the 1st and the 3rd quadrant...
Use the general solution for the tangent: tgx=tga => x=k.pi+a and find out for what values of k the solution is in the given interval.

5. Then you wouldn't want to add 360 degrees, but how much? (This is a multi-faceted question, because the tangent function is periodic with what period?)

7. Okay, got it, so adding pi to -0.7853 and 1.3258 will bring the answers?

8. You'll need to add it several times, until you get outside the allowed region.

9. Sure! Just check the obtained x is in the given iterval.

10. So essentially this:

x=-0.785399+π, x=-0.785399+2π, x=-0.785399+3π; x=1.32582, x=1.32582+π, x=1.32582+2π