Sin theta= sin theta tan theta

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SUMMARY

The equation sin θ = sin θ tan θ can be solved by rewriting it as sin θ - sin θ tan θ = 0. Dividing both sides by sin θ is incorrect as it results in the loss of potential solutions. The correct approach involves factoring the left side to find the values of θ within the domain of 0 to 2π. The solutions occur in the first and third quadrants where tan θ equals 1.

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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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