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

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The equation tan^2x - 1 = 0 simplifies to tan x = ±1, leading to solutions at angles such as π/4, 3π/4, and 5π/4. The general solution can be expressed as π/4 + kπ, where k is an integer, to account for all angles where the tangent equals 1 or -1. The discussion highlights the importance of understanding the periodic nature of trigonometric functions and how to express solutions without listing them all. For other equations, such as sin x = ±1/2, the solutions can vary, requiring different approaches to express the general solutions. Overall, the key takeaway is the method of representing solutions using integer multiples of the fundamental angles.
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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?
 
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Definitely ! Thanks.
 
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