Thread: 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?

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

Radians - I double checked

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.

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?)

If radian, then add pi instead of 180 degrees...

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

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

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

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

Looks correct to me. Always double-check your answers, though.

So 6* answers is typical of tan? Thanks so much I just want to be prepared for my test...

Of course it is!
In this special case it has 8. For anothrer equations it may have less or more.

For that region of interest, you got 6 answers. That would be typical of your original equation that is quadratic in tan(x), if the roots are distinct and real.