# Homework Help: Solving csc x + 2 for 0

1. Mar 27, 2008

### iBankingFTW

[SOLVED] Solving csc x + 2 for 0

1. The problem statement, all variables and given/known data

"Solve csc x + 2 = 0 for 0 <= x < 2 $$\pi$$

Choices are:

A. $$\pi$$/6 and 5$$\pi$$/6
B. $$\pi$$/6 and 7$$\pi$$/6
C. 4$$\pi$$/3 and 5$$\pi$$/3
D. 7$$\pi$$/6 and 11$$\pi$$/6

2. Relevant equations

csc x =2
sin x = 1/2 = 30 degrees = $$\pi$$/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 $$\pi$$/6

So the answer is either A or B but I don't understand where the second answer comes from.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 27, 2008

### rocomath

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: Mar 27, 2008
3. Mar 27, 2008

### iBankingFTW

Ohhhh, ok. So $$\pi$$/6 is in Quadrant I so that's correct and then 5$$\pi$$/6 is in Quad. 2 so that's also correct but 7$$\pi$$/6 is in Quad. 3 which is negative.

4. Mar 27, 2008

### rocomath

No. Where is Sine negative? Definitely not Quad 2 & 3.

5. Mar 27, 2008

### iBankingFTW

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$$\pi$$/6 is in the third quadrant and isn't that negative if it's sine?

Last edited: Mar 27, 2008
6. Mar 27, 2008

### rocomath

Did you fix your first step?

You're not solving for $$\csc x=2$$ ... it's $$\csc x=-2$$

7. Mar 27, 2008

### iBankingFTW

Oh jeez, I'm such an idiot. I hate making little mistakes like that. So the answer is B. then since 7$$\pi$$/6 is in Quadrant 3?

8. Mar 27, 2008

### rocomath

Where else? One more solution!

9. Mar 27, 2008

### iBankingFTW

Haha...thanks for the help.