urgent: sin2x = sin(x-4)

• crimson soldier

  20 Oct 08, 09:06

  I am tertiary engine student...can do double and triple integrals but I can't do this lol =p

   

  one of my friends' tutees encountered this for AMATH for which o levels will be on wednesday.

   

  any bros can help please?  sin2x =sin(x-4) , 0<= x <= 5.

   

  thanks in advance

• eagle

  20 Oct 08, 09:53

  sin 2x = sin (x - 4)
  2x = x-4, π - (x-4), 2π + (x-4), 3π - (x-4), 4π + (x-4), 5π - (x-4) => Generate round the quadrant
  x = -4, (π+4)/3, 2π-4, (3π+4)/3, 4π-4, (5π+4)/3
  x = -4, 2.38, 2.28, 4.47, 8.57, 6.57

  since 0 ≤ x ≤ 5,
  x = 2.28, 2.38, 4.47 (do check for careless mistakes)

   

  Regards,
  Eagle (ExamWorld)

• caleb_chiang

  20 Oct 08, 10:02

  

  I do it the easy way, which must be wrong...

  sin2x =sin(x-4)

  Take away sin

  2x = x - 4

  x = -4;

  

• eagle

  20 Oct 08, 10:32

  Originally posted by caleb_chiang:

  

  I do it the easy way, which must be wrong...

  sin2x =sin(x-4)

  Take away sin

  2x = x - 4

  x = -2;

  

  first wrong
  if 2x = x - 4,
  x = -4

  second wrong,
  0 ≤ x ≤ 5

• caleb_chiang

  20 Oct 08, 10:53

  Originally posted by eagle:

  first wrong
  if 2x = x - 4,
  x = -4

  second wrong,
  0 ≤ x ≤ 5

  
  oh ya...

• wishboy

  20 Oct 08, 10:54

  got another method ! (if im not wrong)

  sin 2x = sin (x - 4)
  sin 2x - sin (x - 4) = 0

  using factor formula,

  cos [(3x - 4)/2] sin [(x + 4)/2] = 0
  cos [(3x - 4)/2] = 0    or    sin [(x + 4)/2] = 0

  ok then from here can continue urself liao

• eagle

  20 Oct 08, 10:56

  Originally posted by wishboy:

  got another method ! (if im not wrong)

  sin 2x = sin (x - 4)
  sin 2x - sin (x - 4) = 0

  using factor formula,

  cos [(3x - 4)/2] sin [(x + 4)/2] = 0
  cos [(3x - 4)/2] = 0    or    sin [(x + 4)/2] = 0

  ok then from here can continue urself liao

  It's a longer method

  because in the end, you will still have to generate the values using the quadrant

• crimson soldier

  20 Oct 08, 11:08

  Originally posted by eagle:

  sin 2x = sin (x - 4)
  2x = x-4, π - (x-4), 2π + (x-4), 3π - (x-4), 4π + (x-4), 5π - (x-4) => Generate round the quadrant
  x = -4, (π+4)/3, 2π-4, (3π+4)/3, 4π-4, (5π+4)/3
  x = -4, 2.38, 2.28, 4.47, 8.57, 6.57

  since 0 ≤ x ≤ 5,
  x = 2.28, 2.38, 4.47 (do check for careless mistakes)

   

  Regards,
  Eagle (ExamWorld)

   

  Thanks for the prompt reply...Anyway sketching out the curve reveals that there is a 4th solution at about x= 0.2 to 0.3. Any help again?

• Lee012lee

  20 Oct 08, 11:10

  Originally posted by caleb_chiang:

  

  I do it the easy way, which must be wrong...

  sin2x =sin(x-4)

  Take away sin

  2x = x - 4

  x = -4;

  

  64/16 = ?

  Cancel the 6's, so

  64/16 = 4/1 = 4

  You might say, using a calculator, 64/16 = 4. Correct.

  Is it really correct ?

• eagle

  20 Oct 08, 11:29

  Originally posted by crimson soldier:

  Thanks for the prompt reply...Anyway sketching out the curve reveals that there is a 4th solution at about x= 0.2 to 0.3. Any help again?

  oh ya, forgot to generate the quadrant backwards

  before 2x=(x-4),

  we also have 2x = -π - (x-4)
  so 3x = 4 - π
  x = 0.286

• syncopation_music

  20 Oct 08, 22:25

  Originally posted by crimson soldier:

  I am tertiary engine student...can do double and triple integrals but I can't do this lol =p

   

  one of my friends' tutees encountered this for AMATH for which o levels will be on wednesday.

   

  any bros can help please?  sin2x =sin(x-4) , 0<= x <= 5.

   

  thanks in advance

  haha...hi crimson.. just in case if u cant rmb me ..i am bohi ...

  Hope u have found ur answer in homework forum

   

  regards