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Solving csc x + 2 for 0

[SOLVED] Solving csc x + 2 for 0

1. 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


2. Homework Equations

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

3. 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.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

1,750
1
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.?
 
1,750
1
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:
1,750
1
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?
 
1,750
1
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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