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How to solve this trig equation in radians 2sin^2(x) + sin(x) - 1=0

  1. Sep 9, 2004 #1
    I just got a review sheet in math that contain questions that I dont know how to do from last year. ONe asks to solve for all values of theta of [0,2pi] in radians: 2sin^2(theta)+sin(theta)-1=0

    any ideas?
  2. jcsd
  3. Sep 9, 2004 #2
    [tex]sin\ 2\theta = 2\ sin\ \theta\ cos\ \theta[/tex]
    [tex]sin^2\ \theta + cos^2\ \theta = 1[/tex]
  4. Sep 9, 2004 #3
    that confuses me more. I have no clue what the question is even asking, I think I forgot all my math. Could you give a longer explanation?
  5. Sep 9, 2004 #4
    Let [tex]sin\ \theta\ = x[/tex]

    Then your original equation now looks like [tex]2x^2 + x - 1 = 0[/tex].

    Solve the quadratic equation. Once you do that, the rest should be easy.
  6. Sep 9, 2004 #5


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    The question is asking for which angles in a 0 to 2pi interval satisfy that equation.

  7. Sep 10, 2004 #6
    forget about what i have said. use recon's suggestion and understand cyclovenom's point.
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