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

  • Thread starter iBankingFTW
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In summary, the equation csc x + 2 = 0 has two solutions, \pi/6 and 7\pi/6, and both of these values lie in the third and fourth quadrants where sine is negative. The correct answer is B.
  • #1
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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  • #2
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:
  • #3
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.?
 
  • #4
No. Where is Sine negative? Definitely not Quad 2 & 3.
 
  • #5
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:
  • #6
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 ...
 
  • #7
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?
 
  • #8
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!
 
  • #9
Haha...thanks for the help. :cry:
 

What does "Solving csc x + 2 for 0" mean?

Solving csc x + 2 for 0 means finding the value of x that makes the expression equal to 0.

Why is it important to solve csc x + 2 for 0?

Solving csc x + 2 for 0 can help to find the roots or solutions of a trigonometric equation, which can be used in various applications in fields such as engineering, physics, and mathematics.

How do you solve csc x + 2 for 0?

To solve csc x + 2 for 0, you can use algebraic manipulations and trigonometric identities to isolate the variable x and find its value. It may also involve using a calculator or graphing software to approximate the solution.

What are the possible solutions for csc x + 2 = 0?

The possible solutions for csc x + 2 = 0 are values of x that make the expression equal to 0. This can include both real and complex solutions, depending on the specific equation and domain.

Can you provide an example of solving csc x + 2 for 0?

Sure, for the equation csc x + 2 = 0, we can use the trigonometric identity csc x = 1/sin x to rewrite it as 1/sin x + 2 = 0. Then, we can multiply both sides by sin x to get 1 + 2sin x = 0. Solving for sin x, we get -1/2. Using the inverse sine function, we can find two solutions for x: x = -π/6 and x = -5π/6. These are the values of x that make csc x + 2 equal to 0.

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