Solving tan^2 x + 6tanx - 7 Algebraically over 0 and 2π

[seiferseph]

solve algebraically over 0 and 2pi. then give general solution

tan^2 x + 6tanx - 7

so i factored it and got

(tanx +7)(tanx - 1)
tanx = -7, 1

then do you reject the -7? why? i graphed it and only saw the two solutions from tanx = 1

 

[whozum]

$$x = arctan(-7), x = arctan(1)$$

They both exist. Check if theyre in your domain.

 

[Knavish]

You reject x=arctan(-7) because the answer isn't in the domain of 0 to 2pi. The other answer for this is arctan(1) + pi.

 

[seiferseph]

You reject x=arctan(-7) because the answer isn't in the domain of 0 to 2pi. The other answer for this is arctan(1) + pi.

ok i get it, thanks! so you solve to get x = pi/4 + n(pi)
for a general solution, right?

 

[Knavish]

Actually, it's just x = [answer] + n(pi) for tan(x); this is because tan(x) is equal at intervals of pi (or 180 degrees).

 

[Curious3141]

There's no reason why the solution for tan x = -7 should be rejected. tan(x) is periodic with period of pi, and all you have to do is add multiples of pi to the calculator value to get it into the required range.

If you plug in arctan(-7) into a calculator in radians mode, you'll get -1.429 rad. Just add pi to it, and you have a valid solution. Add pi another time, and you have another.

 

[Knavish]

Sorry, I didn't exactly mean "rejected" as "invalid." I meant we couldn't use the solution as an answer to the problem.

And, yeah, the other answers to the problem are arctan(-7)+n(pi), where n=1 and n=2.

 

[seiferseph]

i was confused by that because when i graphed it i only got the two solutions, so there are 4 solutions?