Sin theta= sin theta tan theta

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The equation sin(theta) = sin(theta) tan(theta) can be solved by factoring rather than dividing by sin(theta), which risks losing solutions. The correct approach is to rewrite the equation as sin(theta) - sin(theta) tan(theta) = 0. This allows for factoring out sin(theta), leading to the solutions for theta. The discussion emphasizes the importance of maintaining all potential solutions when manipulating trigonometric equations. Understanding this method is crucial for accurate problem-solving in trigonometry.
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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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