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Trig equations, solve for x?

  1. Jul 27, 2010 #1
    I had some questions on a test and i got these two wrong..

    x is greater than or equal to 0 and x is less than or equal to 360

    1) 2sin x + sin x cos x = 0
    2) sin 2x = -1/sqrt(2)
     
  2. jcsd
  3. Jul 28, 2010 #2

    Mark44

    Staff: Mentor

    Show us what you did and we'll go from there...
     
  4. Jul 28, 2010 #3
    2Sin x + cos x sin x = 0
    Sin x(2 + cos x) = 0
    sin x = 0
    x= sin^-1(0)
    x=0degrees
    or
    2+ cos x= 0
    cos x =-2
    x= cos^-1(-2)
    x=?

    I did something different on the test but i think this is wrong too >.<

    Don't make show what I did for #2 :(
     
  5. Jul 28, 2010 #4
    Well, cos x = 2 does not have any solutions in x. This is because the range of cos x is [-1, 1]
     
  6. Jul 28, 2010 #5

    Mark44

    Staff: Mentor

    This is one solutions; there are two other numbers x in [0, 360] (deg.) for which sin(x) = 0. The inverse sine function will only produce numbers in the interval [-90, 90], in degrees. Think about where the graph of y = sin(x) crosses the x-axis.
    OK, maybe I can help you get started in the right direction. If sin(2x) = -1/sqrt(2), then 2x = sin-1(-1/sqrt(2)). What are the possible values of 2x in [0, 720] (deg), so that sin(2x) = -1/sqrt(2)?
     
  7. Jul 28, 2010 #6
    it's long time that I haven't posted anything here and so I forgot everything about latex. Will anybody here consider giving me link so that I can learn to type it.

    then I'll post solution for this problem.
     
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