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Trig Problem- solve equation

  1. Jan 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the equation 2 - cos2x =sinx. Give the solution in the interval 0≤x≤360.

    2. Relevant equations
    sin2x+cos2x=1
    I know for sin you take inverse. Then subtract that from 180. I believe those are basic angle then you +/- 360. ?

    3. The attempt at a solution

    I tried to manipulate it to work and solve but im not sure what else and then how to.
    2 - cos2x =sinx
    2(1-cos2x)=sinx
    2(sin2x)=sinx
     
    Last edited: Jan 2, 2012
  2. jcsd
  3. Jan 2, 2012 #2
    Its too late at night for me to start working that out but have you tried the trig identity

    tanx=sinx/cosx
     
  4. Jan 2, 2012 #3
    Um, i think so. Wait yes i did.
     
  5. Jan 2, 2012 #4
    I would suggest writing it in a form that's quadratic in sin(x). Use the squared identity but in the other way.

    edit - also, this step:[itex]2(1-cos^{2}(x))=sin(x)[/itex] is NOT valid. Can you see why?
     
    Last edited: Jan 2, 2012
  6. Jan 2, 2012 #5

    SammyS

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    The following two equations are not equivalent.

     2 - cos2x = sinx    &     2(1-cos2x) = sinx

    The first equation is equivalent to
    1 + (1 - cos2x) = sinx​

    Now substitute sin2x for 1 - cos2x and then subtract sinx from both sides. You then have a quadratic equation in sinx .
     
  7. Jan 2, 2012 #6

    Curious3141

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    Of course, the quadratic may not have real roots (and doesn't in this case). This is a pickle if you want real solutions for x.

    (To be less cryptic, I meant that there are no real solutions here).
     
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