What are the solutions to tan^2x - 1 = 0?

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Homework Help Overview

The problem involves solving the equation tan²x - 1 = 0, which relates to trigonometric functions and their periodic nature.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of solving the equation, including taking square roots and identifying angles where the tangent function equals ±1. There are questions about how to express the general solution and the implications of different values of k.

Discussion Status

Participants are exploring various ways to express the solutions, including the use of k to denote integer multiples for periodic angles. There is an ongoing dialogue about the differences between specific angles and how to generalize solutions for other trigonometric equations.

Contextual Notes

Some participants question how to list all solutions without explicitly enumerating them, and there is discussion about the periodic nature of trigonometric functions, including the differences in solutions for sine and tangent equations.

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Homework Statement



Tan^{2}X - 1 = 0

Homework Equations





The Attempt at a Solution



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

tan x = \pm1

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

Do I just write pi/4 + 2kpi as my answer?
 
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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.
 
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How do I list every spot where tan = 1 or -1? without actually "listing" them all out.
 
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?
 
OH, so for this specific problem I'm going to write:

pi/4, 3pi/4, 5pi/4, 7pi/4, + pi/2? In that order?
 
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.
 
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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.
 
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:
Definitely ! Thanks.
 

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