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Inverse trig.

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    I have two questions sin2x = 1/25 and this obviously becomes sinx= +-(1/5)
    I also have cos2-1.5cosx-0.54 and cosx = (-3/10) and (9/5)

    Now this is asking for me to solve for the x value in radians in the domain [0,2pi] and I have no idea how to solve these for exact values. Help would be appreciated.
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Sep 22, 2013 #2

    Mark44

    Staff: Mentor

    What does "cos2-1.5cosx-0.54 " mean?
    cos(x) can't possibly equal 9/5.
     
  4. Sep 22, 2013 #3
    that was my second equation, and my bad for putting it down as it is an extraneous root.
     
  5. Sep 22, 2013 #4

    Mark44

    Staff: Mentor

    "cos2-1.5cosx-0.54" is NOT an equation.

    Did you mean cos2(x) - 1.5cosx-0.54 = 0?

    Are there two separate questions, or do you have a question about a system of two equations?

    Help us out here - don't make us guess about this stuff...
     
  6. Sep 22, 2013 #5
    Sorry, it is two questions. And yes those are my two problems that I need to solve over the domain [0,2pi]
    I need help solving for x using inverse trig.
     
  7. Sep 22, 2013 #6

    Mark44

    Staff: Mentor

    For the first question, if you had sin(x) = ±1/4, there are two numbers in [0, ##2\pi##] for which sin(x) = 1/4 and two more in this interval for which sin(x) = -1/4.

    If we let θ be the smallest of the four values, we have sin(θ) = 1/4, so θ = sin-1(1/4). The other value in that interval for which sin(θ) = 1/4 is ##\pi - \theta##, or ##\pi - sin^{-1}(1/4)##. These are the exact values.

    Similar work will get you the two values for which sin(θ) = -1/4.

    Your first problem is similar to this.
     
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