Problem Check

1. Jul 8, 2009

I'm

1. The problem statement, all variables and given/known data

Tan$$^{2}$$X - 1 = 0

2. Relevant equations

3. The attempt at a solution

I added 1 to both sides, then I took the square root of both sides giving me:

tan x = $$\pm$$1

which gives me answers at pi/4, 3pi/4, 5pi/4 and so on.

Do I just write pi/4 + 2kpi as my answer?

2. Jul 8, 2009

Cyosis

All the answers you've listed are correct, however depending on what k is incomplete. I assume k is an integer. If so then your final answer are points that are all on the same 'spot' of the unit circle. There are more spots where the tangent is 1 or -1.

Last edited: Jul 8, 2009
3. Jul 8, 2009

I'm

How do I list every spot where tan = 1 or -1? without actually "listing" them all out.

4. Jul 8, 2009

Cyosis

What you're doing now is making 2Pi rotations, you could also make pi rotations or pi/2 rotations or pi/3 rotations etc. You need to find the angle that fits your equation.

What is the difference between pi/4, 3pi/4, 5pi/4, etc?

5. Jul 8, 2009

I'm

OH, so for this specific problem I'm going to write:

pi/4, 3pi/4, 5pi/4, 7pi/4, + pi/2? In that order?

6. Jul 8, 2009

Bohrok

No, you can start with pi/4 + something with k where, for every integer k, you will get pi/4, 3pi/4, 5pi/4, 7pi/4, etc. as well as the negative solutions -pi/4, -3pi/4, -5pi/4, -7pi/4, etc.

Last edited: Jul 8, 2009
7. Jul 8, 2009

I'm

SO my final answer will be:

pi/4 + 2pi/4?

How would I do this for a repeating angle such as:

sin x = -1/2 or 1/2 with no limits? since the difference are different.

Like pi/6, 5pi/6, 7pi/6.

The differences are different.

8. Jul 8, 2009

Bohrok

As Cyosis said, What is the difference between pi/4, 3pi/4, 5pi/4, etc?
The solution will be pi/4 + k*(common difference between each of the numbers above)

For a problem like sin2x = 1/4 or sin x = -1/2, 1/2 where the differences are different, you give "two" solutions
x = pi/6 + 2k*pi and x = 5pi/6 + 2k*pi
or sometimes as
x = pi/6 + 2k*pi, 5pi/6 + 2k*pi

Make sense?

Last edited: Jul 8, 2009
9. Jul 8, 2009

I'm

Definitely ! Thanks.