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Solve cot/cosec

  1. Jan 31, 2010 #1
    1. Solve 26cot2Theta=cosec2Theta for -π≤0≤π



    2. Relevant equations



    3. Well. I tried to solve this and ended up with 26cos2Theta=0. That can't be right.
    I started with 26(cos2Theta/sin2)Theta=1/sin2Theta


    Guess that was a wrong start. Any suggestions please?
     
  2. jcsd
  3. Jan 31, 2010 #2

    Mark44

    Staff: Mentor

    Multiply both sides by sin2(theta). When you multiply the right side by sin2(theta) you don't get 0.
     
  4. Jan 31, 2010 #3
    26cos2Thetasin2Theta=sin2Theta
    = 26cos2Theta

    really?
     
  5. Jan 31, 2010 #4

    Mark44

    Staff: Mentor

    No, that's not it. cot(theta) = cos(theta)/sin(theta), right? And csc(theta) = 1/sin(theta). Those are the identities to use.
     
  6. Jan 31, 2010 #5
    so cot2Theta=(cos2Theta/sin2Theta)?
     
  7. Jan 31, 2010 #6

    Mark44

    Staff: Mentor

    Yes, and csc2(theta) = 1/sin2(theta)
     
  8. Feb 3, 2010 #7
    How about this?
    26cot2x=1+cot2x
    25cot2x-1=0
    25/tan2x-1=0
    25-tan2x=tan2x
    2tan2x-25=0

    can I go with that?
     
  9. Feb 3, 2010 #8

    Mark44

    Staff: Mentor

    In the next step after the line above, you should add 1 to both sides. After that, multiply both sides by tan2x.

    The line below is wrong, which makes the line after it wrong, too. How did you get from the line above to this one?
    No.
     
  10. Feb 3, 2010 #9
    I got to the next line by multiplying out tanx2 from the division.
    Why would I add 1 to both sides. It will get rid of the -1 and put 1 to the right of the equation?
     
  11. Feb 3, 2010 #10
    That would give:
    26cot2x=1+cot2x
    25cot2x-1=0
    (25/tan2x)-1=0
    (25/tan2x)=1
    right?
     
  12. Feb 3, 2010 #11

    Mark44

    Staff: Mentor

    25/tan2(x) - 1 = 0
    ==> 25/tan2(x) = 1
    ==> 25 = tan2(x)
    So tan(x) = ?

    Doing it the way you did it,
    25/tan2(x) - 1 = 0
    ==> 25 - tan2(x) = 0 Note that 0*tan2(x) is not tan2(x).
    Now you want to add tan2(x) to both sides, which gets you to my last line above.
     
  13. Feb 3, 2010 #12

    Mark44

    Staff: Mentor

    This would be confusing to a casual reader. All of the things you have as cot2x and tan2x are really cot(x) and tan2(x).
     
  14. Feb 3, 2010 #13
    ahh wait - sorry
    I'm having issues with the editing software here.
     
  15. Feb 3, 2010 #14
    That would give:
    26cot2x=1+cot2x
    25cot2x-1=0
    (25/tan2x)-1=0
    (25/tan2x)=1
    right?
     
  16. Feb 3, 2010 #15
    tanx=5
    Thanks Mark44
     
  17. Feb 3, 2010 #16
    so my solutions should be at 1.37 (2d.p.)
    and -π+1.37 = -1.77 (2d.p.)
     
  18. Feb 3, 2010 #17

    Mark44

    Staff: Mentor

    That's one solution. What's the other one?
     
  19. Feb 3, 2010 #18
    err.. don't follow :confused:

    I gave two solutions
    solutions should be at 1.37 (2d.p.)
    and -π+1.37 = -1.77 (2d.p.)
     
  20. Feb 3, 2010 #19

    Mark44

    Staff: Mentor

    The equation tan2(x) = 25 has two solutions in terms of tan(x), and you got one of them, which led to two values of x.

    There are two more values of x in the interval -pi <= x <= pi, meaning there are four solutions all together.
     
  21. Feb 4, 2010 #20
    ahh! Tanx=+or -5
    So the other two solutions are at -1.37 and 1.77 (2d.p.) :tongue2:
     
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