Math Problems

• Chockeh wrote on 2012-02-11 14:34
  So I solved most of my webwork except for three problems.
  
  1)
  
  

  [Image: http://puu.sh/guiE]

  
  
  2)
  
  

  [Image: http://puu.sh/guiF]

  
  
  3)
  
  

  [Image: http://puu.sh/guiH]

  
  
  You may (or may not) need this to solve:
  
  

  [Image: http://www.mathsisfun.com/geometry/images/circle-unit-radians.gif]

  
  It would also be nice to have the steps needed to solve the problems.
• Senbonzakura wrote on 2012-02-11 14:45
  [Image: http://puu.sh/guiEv]

  1.
  
  

  [Image: http://puu.sh/guiE]

  
  
  Solution:
  
  

  [Image: http://img217.imageshack.us/img217/1244/53510382.jpg]
• Phunkie wrote on 2012-02-11 14:57
  Tried to help you out, but it's too late for me to be doing this kind of math.
  
  Try using Wolfram Alpha.
  
  It's tons of help.
• pkMaster wrote on 2012-02-11 15:50
  I'll skip question 1 since it's already done. I don't have the exact values of x for each problem since my calculator is dead, but you should be able to figure it out on your own.
  
  Solution:
  [SPOILER="Spoiler"]For question 2:
  3sin^2(x) - 7sin(x) + 2 = 0
  (sin(x) - 2) (3sin(x) - 1) = 0; The first solultion yeilds no results
  3sin(x) = 1
  sin(x) = 1/3
  
  For question 3:
  sin^2(x) = 1/9; Find the square root of both sides
  sin(x) = 1/3 and -1/3
  [/SPOILER]
  
  Answer:
  [SPOILER="Spoiler"]For question 2:
  x = sin^-1( 1/3 )
  Not sure what the exact value is since my calc is dead.
  
  For question 3:
  x = sin^-1(1/3) and sin^-1(-1/3)[/SPOILER]
• Senbonzakura wrote on 2012-02-11 15:59
  The CAST rule is useful for these.
  
  Sine is positive in quadrants 1 and 2, negative in 3 and 4.
  
  sin x = 1/3
  x = 13 pi/120 and 107 pi/120.
  
  sin x = -1/3
  x = 133 pi/120 and 227 pi/120.
  
  These are just approximations.
• Kenny? wrote on 2012-02-11 20:26
  My camera sucks so :l
  1. pi/3 or 4pi/3
  2 and 3, can't find exact pi values, so I just did sininverse(1/3)
  [SPOILER="Spoiler"]

  [Image: http://i.minus.com/ixDpw8CzbAjv4.JPG]

  [/SPOILER]
  [SPOILER="Spoiler"]

  [Image: http://mabination.com/57837cd8-399e-4c47-8fd0-515fb54b7333]

  [Image: http://i.minus.com/irCpksY34LbIs.JPG]

  [/SPOILER]
  

  [Image: http://mabination.com/57837cd8-399e-4c47-8fd0-515fb54b7333]
• Groom wrote on 2012-02-11 23:19
  [spoiler]

  [Image: http://img813.imageshack.us/img813/7696/mittrigonometrygraphs.jpg]

  [/spoiler]
  
  Please see the attached MIT trigonometry graph
  
  1.
  2s^2 -5c - 4
  2(1-c^2) -5c -4
  2 - 2c^2 - 5c - 4
  2c^2 +5c +2
  (2c+1)(c+2)
  c= -1/2, -2, but x=cos^-1 (-2) can't be a value
  x = 2Ï€/3, 4Ï€/3
  
  2.
  3s^2 -7s +2
  (3s-1)(s-2)
  s= 1/3, 2, but x=sin^-1 (2) can't be a value
  x1= 0.34, x2= π - 0.34 = 2.80 approximately
  
  3.
  s^2 = 1/9
  s = 1/3, -1/3
  x = 0.34, 2.8 from q2
  and π + 0.34 = 3.48, 2π - 0.34 = 5.94 approx.
• Chockeh wrote on 2012-02-12 02:48
  Thanks a lot guys.