# Trig equation

1. Jun 23, 2005

### 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

2. Jun 23, 2005

### whozum

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

They both exist. Check if theyre in your domain.

3. Jun 23, 2005

### 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.

4. Jun 23, 2005

### seiferseph

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

5. Jun 23, 2005

### 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).

6. Jun 23, 2005

### 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.

7. Jun 23, 2005

### 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.

Last edited: Jun 23, 2005
8. Jun 24, 2005

### seiferseph

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