What are the solutions to csc x + 2 = 0 for 0 <= x < 2 \pi?

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Homework Help Overview

The problem involves solving the equation csc x + 2 = 0 within the interval 0 <= x < 2π. Participants are exploring the implications of the cosecant function and its relationship to the sine function.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the algebraic manipulation of the equation and the implications of the cosecant function's values. There is confusion regarding the correct quadrants where sine is negative and how that affects the solutions.

Discussion Status

The discussion is ongoing, with participants questioning their understanding of the algebra involved and the properties of the sine function in different quadrants. Some guidance has been offered regarding the correct interpretation of the equation and the quadrants to consider.

Contextual Notes

There is a noted misunderstanding regarding the sign of the cosecant function and its relation to the sine function, leading to confusion about the possible solutions. Participants are also referencing external knowledge and personal discussions to clarify their reasoning.

iBankingFTW
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[SOLVED] Solving csc x + 2 for 0

Homework Statement



"Solve csc x + 2 = 0 for 0 <= x < 2 [tex]\pi[/tex]

Choices are:

A. [tex]\pi[/tex]/6 and 5[tex]\pi[/tex]/6
B. [tex]\pi[/tex]/6 and 7[tex]\pi[/tex]/6
C. 4[tex]\pi[/tex]/3 and 5[tex]\pi[/tex]/3
D. 7[tex]\pi[/tex]/6 and 11[tex]\pi[/tex]/6


Homework Equations



csc x =2
sin x = 1/2 = 30 degrees = [tex]\pi[/tex]/6

The Attempt at a Solution



From csc x + 2 = 0

I get csc x = 2

Which is:

1/sin = 1/2 and I know that 1/2 is [tex]\pi[/tex]/6


So the answer is either A or B but I don't understand where the second answer comes from.
 
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You messed up your Algebra. You will get 2 solutions b/c Cosecant takes on that value at 2 places. Sine is negative in what Quadrants? Thus, Cosecant is also negative at those 2 places.
 
Last edited:
Ohhhh, ok. So [tex]\pi[/tex]/6 is in Quadrant I so that's correct and then 5[tex]\pi[/tex]/6 is in Quad. 2 so that's also correct but 7[tex]\pi[/tex]/6 is in Quad. 3 which is negative.

So the answer is A.?
 
No. Where is Sine negative? Definitely not Quad 2 & 3.
 
Tangent is positive in Quadrant 3 (ASTC as I learned it), so wouldn't sine be negative there?

I asked my dad about this too and he said that the answer is B. but I don't understand because 7[tex]\pi[/tex]/6 is in the third quadrant and isn't that negative if it's sine?
 
Last edited:
Did you fix your first step?

You're not solving for [tex]\csc x=2[/tex] ... it's [tex]\csc x=-2[/tex]

Check your Algebra again! So your values should be in Quadrants 3 & 4 ...
 
Oh jeez, I'm such an idiot. I hate making little mistakes like that. So the answer is B. then since 7[tex]\pi[/tex]/6 is in Quadrant 3?
 
iBankingFTW said:
Oh jeez, I'm such an idiot. I hate making little mistakes like that. So the answer is B. then since 7[tex]\pi[/tex]/6 is in Quadrant 3?
Where else? One more solution!
 
Haha...thanks for the help. :cry:
 

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