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Radian angle solving

  1. May 30, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]2sin(x-\frac{\pi}{3}) = 1[/tex]
    for the range 0 <= x<= 2π

    The attempt at a solution

    so [tex]x-\frac{\pi}{3} = \frac{\pi}{6}[/tex]

    i can get the first solution which is π/2 but how do i get the next solutions?

    Thank you
  2. jcsd
  3. May 30, 2008 #2

    Doc Al

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    Staff: Mentor

    Examine the graph of a sine function between 0 and 2π. Find all angles where sine = 1/2. π/6 is only one of them.
  4. May 30, 2008 #3
    well there is 1 more in the range of 0 to π to 360

    that is π - π/6
    = 5π/6

    so the answer is

    π and 5π/6

    is there any more? i presume i dont count the -1/2s

    thanks :)
  5. May 30, 2008 #4

    Doc Al

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    Staff: Mentor

    Right. So the two values of [itex]\theta[/itex] that satisfy [itex]\sin\theta = 1/2[/itex] are [itex]\pi/6[/itex] & [itex]5\pi/6[/itex].

    Given that, what are the values for x?
  6. May 30, 2008 #5
    do you add π/3 to both of them?

  7. May 30, 2008 #6

    Doc Al

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    Staff: Mentor

  8. Jun 1, 2008 #7
    This is the way I was taught to do it :
    [tex]2\sin (x-\frac {\pi}{3})= 1[/tex]
    [tex]\sin (x-\frac {\pi}{3})= \frac {1} {2}[/tex]
    Let [tex]q =x-\frac {\pi}{3} [/tex]
    [tex]\sin (q)= \frac {1} {2}[/tex]
    Where is the [tex]\sin q = \frac {1}{2}[/tex] ?
    At [tex]\frac {\pi}{6}, \frac {5\pi }{6}[/tex]
    Thus : [tex]q_{1} =\frac {\pi}{6} , q_{2}=\frac {5\pi }{6}[/tex]
    We're not done. We still have to solve the x. Note that I was supposed to add 2 Pi to q 1 and q 2, but if you do it separately, you will see the solutions would not be needed since they are outside of 2 Pi when we add pi /3.
    Continuing : Simply setting the q's equal to x - pi /3
    [tex]x-\frac {\pi}{3} =\frac {\pi}{6}[/tex]
    [tex]x-\frac {\pi}{3} =\frac {5\pi }{6}[/tex]
    Solving, we get the solutions to be : [tex]x_{1} = \frac {\pi}{2},x_{2} = \frac {7\pi}{6},[/tex]

    In your calculator, if you graph these two functions, you will see the solutions to be those as noted.
  9. Jun 1, 2008 #8
    Thanks!!!! :)
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