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Solving csc x + 2 for 0

  1. Mar 27, 2008 #1
    [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 [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. Relevant 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. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 27, 2008 #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: Mar 27, 2008
  4. Mar 27, 2008 #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.?
     
  5. Mar 27, 2008 #4
    No. Where is Sine negative? Definitely not Quad 2 & 3.
     
  6. Mar 27, 2008 #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: Mar 27, 2008
  7. Mar 27, 2008 #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 ...
     
  8. Mar 27, 2008 #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?
     
  9. Mar 27, 2008 #8
    Where else? One more solution!
     
  10. Mar 27, 2008 #9
    Haha...thanks for the help. :cry:
     
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