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Homework Help: Trig. again

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data
    this time i have posted two questions on a post, instead of me creating posts all over the forum.

    a)cot^2x + cosec^2x = 7
    b) (2 cosec^2)2x + cot2x =3
    find out the solutions.
    2. Relevant equations
    cosec^2 = 1+ cot^2


    3. The attempt at a solution
    a) cot^2x + cosec^2x = 7
    cosec^2-1 + cosec^2 = 7
    2cosec^2x = 8
    now i thought of solving it by the difference of squares?
    couldn't solve it =/
    b) totally cluesless as this one has 2 after function cot i.e cot2x =/
     
  2. jcsd
  3. Dec 11, 2009 #2

    Mark44

    Staff: Mentor

    It's not very clear what the equation above is saying. You are at least using parentheses, but not in the most helpful way.
    Is this supposed to be 2csc^2(2x) + cot(2x) = 3?

    If so, you should replace csc(2x) with 1/sin(2x) and cot(2x) with cos(2x)/sin(2x).
    So csc^2(2x) = 4
    Or 1/[sin^2(2x)] = 4
    Can you take it from there?
     
  4. Dec 11, 2009 #3
    sorry as i am pretty newbie with latex =/ hmm i dont get it, how did you end up with = 4? and yea i gotta say.. that question doesn't make sense well the way i typed makes it more complicated (looking)
     
  5. Dec 11, 2009 #4

    Mark44

    Staff: Mentor

    Divide both sides of the equation by 2 to get csc^2(2x) = 4.
     
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