# 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 $$\pi$$

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. Homework 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. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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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.

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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.

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

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Did you fix your first step?

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

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?
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?
Haha...thanks for the help. 