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  • 03-10-2013, 11:02 PM
    Defy
    young adult :)
  • 03-10-2013, 11:31 PM
    KaosPenguin
    Old enough to know better, young enough not to care.
  • 03-11-2013, 12:32 AM
    Dlusion
    Let sin 2x = k

    k = 22- 8√7

    k - 22 = -8√7 [Bring 22 to the LHS)

    (k-22)² = (-8√7)² [Squaring both sides]

    k² - 44k + 484 = 448 [Expanding]

    k² - 44k = 448-484 [Bringing 484 to the RHS]

    k² - 44k = -36 [Multiply throughout by k]

    k³ - 44k² = -36k

    k³ - 44k² + 36k = 0 [Bringing -36k to the LHS]

    k³ - 49k² +5k² + 8k + 28k = 0 [Rearranging factors]

    k³ + 5k² + 8k = 49k² - 28k [Rearranging factors and bring 49k² - 28k to the RHS]

    k³ + 4k² + k² + 4k + 4k = 49k² - 28k [Rearranging factors]

    k³ + 4k² + k² + 4k + 4k + 4 - 4 = 49k² - 28k [Adding + 4 and -4 to make it a quadratic/polynomial]

    k³ + 4k² + k² + 4k + 4k + 4 = 49k² - 28k + 4 [Bringing -4 to the RHS]

    k³ + 4k² + k² + 4k + 4k + 4 = (7k - 2)² [Writing 49k² - 28k + 4 in the form of (a-b)²]

    (1+k)(k² + 4k + 4) = (7k - 2)² [Factoring LHS]

    (1+k)(k + 2)² = (7k - 2)² [Writing k² + 4k + 4 in the form of (a+b)²]

    (1+k) = (7k-2)²/(k+2)² [Bringing (k+2)² to the RHS]

    (1 + 2sinxcosx) = (7k-2)²/(k+2)² [k = sin2x = 2sinxcosx]

    (sin²x + cos²x + 2sinxcosx) = (7k-2)²/(k+2)² [ 1 = sin²x + cos²x and we bring LHS to the form (a+b)²]

    (sinx + cosx)² = (7k-2)²/(k+2)²

    (sinx + cosx) = (7k-2)/(k+2) [Taking root on both sides]

    sinx + cosx = (7k -2)/k / (k+2)/k) [Divide Numerator and Denominator of RHS by k]

    sinx + cosx = (7 - 2/k) / (1+2/k)

    (sinx + cosx) (1 + 2/k) = 7 - 2/k [Taking (1+2/k) to the LHS]

    sinx + cosx + 2sinx/k + 2cosx/k = 7 - 2/k [Multiplying and Expanding]

    sinx + cosx + 2/k(sinx + cosx) + 2/k = 7 [Taking Common factor out]

    sinx + cosx + 2/k(sinx + cosx + 1) = 7 [Taking Common factor out]

    sinx + cosx + 2/k(sinx + cosx + sin²x + cos²x) = 7 [ 1 = sin²x + cos²x]

    sinx + cosx + 2/2sinxcos (sinx + cosx + sin²x + cos²x) = 7 [k = sin2x = 2sinxcosx]

    sinx + cosx + 1/cosx + 1/sinx + sinx/cosx + cosx/sinx = 7 [Canceling common terms]

    sinx+cosx+secx+cosecx+tanx+cotx = 7 [1/cosx=secx,1/sinx=cosecx,sinx/cosx=tan…

    Therefore sinx + cosx + secx + cosecx + tanx + cotx = 7

    3(sinx + cosx + secx + cosecx + tanx + cotx) = My age.
  • 03-11-2013, 12:43 AM
    Protonix
    So you're 21.

    Child's play.
  • 03-11-2013, 01:08 AM
    A s0t
    Quote:

    Originally Posted by Protonix View Post
    So you're 21.

    Child's play.

    you forgot to divide by Zero
  • 03-11-2013, 01:12 AM
    Protonix
    Nonsense.!
  • 03-11-2013, 01:47 AM
    Dlusion
    Quote:

    Originally Posted by Dlusion View Post
    Therefore sinx + cosx + secx + cosecx + tanx + cotx = 7

    3(sinx + cosx + secx + cosecx + tanx + cotx) = My age.

    Is what makes it child's play =] GG
  • 03-12-2013, 03:13 PM
    Tenz
    Im 26 i think... i honestly dont recall at the moment how strange...


    EDIT: i was right 26 LOL
  • 03-12-2013, 03:16 PM
    Far_Too_Jones
    25 til I'm 30 and I'm 30 in October.
  • 03-12-2013, 03:17 PM
    Paladin
    Quote:

    Originally Posted by Tenz View Post
    Im 26 i think... i honestly dont recall at the moment how strange...


    EDIT: i was right 26 LOL

    Are you a po/thead?
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