Sin theta= sin theta tan theta

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Homework Help Overview

The discussion revolves around solving the equation sin(theta) = sin(theta) tan(theta) within the domain of theta from 0 to 2π. Participants are exploring the implications of manipulating the equation, particularly the effects of dividing by sin(theta>.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt to simplify the equation by dividing both sides by sin(theta), leading to the conclusion that tan(theta) = 1. Others question the validity of this approach, noting that it may result in lost solutions.

Discussion Status

The discussion is active, with participants offering alternative methods for rewriting the equation. There is a recognition of the potential pitfalls in the original approach, and some clarity is emerging regarding the factoring of the equation.

Contextual Notes

Participants are considering the implications of dividing by sin(theta), which may exclude certain solutions. The domain of the problem is specified as theta between 0 and 2π.

Coco12
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Problem statement
Solve each equation over the domain theta greater than or equal to 0, less than or equal to 2 pi:

Sin theta= sin theta tan theta

Revelant equations

Problem statement

I divided it by sin theta to get tan theta equal to 1. Tan theta=1 can occur in the first and third quadrant. So from there I solve accordingly. Is this right?
 
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Coco12 said:
Problem statement
Solve each equation over the domain theta greater than or equal to 0, less than or equal to 2 pi:

Sin theta= sin theta tan theta

Revelant equations

Problem statement

I divided it by sin theta to get tan theta equal to 1. Tan theta=1 can occur in the first and third quadrant. So from there I solve accordingly. Is this right?
What about the sinθ?
 
Coco12 said:
I divided it by sin theta to get tan theta equal to 1. Tan theta=1 can occur in the first and third quadrant. So from there I solve accordingly. Is this right?
No.
When you divide both sides by sinθ, you lose solutions. Instead, rewrite the equation like so -
sinθ - sinθ tanθ = 0
and then factor the left side.
 
Mark44 said:
No.
When you divide both sides by sinθ, you lose solutions. Instead, rewrite the equation like so -
sinθ - sinθ tanθ = 0
and then factor the left side.

Oh... I see it now.
 

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