Trigo questions

• je261

  26 Sep 11, 10:30

  1) 1+ 2cos(3x/2 +75deg) = 0

  Ans: 30, 110, 270, 350

  I only managed to get the first 2, how do i get 270 & 350?

   

  2) sec(x/2 - 107deg) = 2

  Ans: 94, 334

  I only managed to get 334, how do i get 94?

   

  Thanks in advance!

• HyuugaNeji

  26 Sep 11, 11:02

  I give you the range.

  If they want x to be between 0 and 360,

  then 0<x<360

  0<3x/2<540
  75<3x/2 +75<615

  Happy?

• dreamer87

  26 Sep 11, 16:56

  1+ 2cos(3x/2 +75deg) = 0
  cos(3x/2 + 75) = -1/2
  cos(alpha) = -1/2,  0 < alpha < 90
  alpha = 60 deg
  
  cos(alpha) = -1/2 ==> cos is negative ==> 2nd and 3rd quarter
  3x/2 + 75 = 180 - alpha, 180+alpha, 360 + 180 + alpha, 360 + 180 - alpha
  3x/2 + 75 = 120, 240, 600, 480
  3x/2 = 45, 165, 525, 405
  x = 30, 110, 350, 270

  
  sec(x/2 - 107deg) = 2
  1/cos(x/2 - 107deg) = 2
  cos(x/2 - 107deg) = 1/2
  cos(alpha) = 1/2
  alpha = 60

  cos(alpha) = 1/2 ==> first and fourth quarter
  0<x<360 ==> 0<x/2<180 ==> -107 < x/2-107<73
  
  Therefore,
  x/2 - 107deg = -alpha ,alpha,
  x/2 - 107deg = -60, 60
  x/2 = 47,167
  x = 94, 334
  
  Hope it helps!

• Darkness_hacker99

  26 Sep 11, 17:14

  thx for sharing

• je261

  30 Sep 11, 20:10

  Thanks dreamer87, your detailed working has helped alot!