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Trig quadratic equation

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    2 cos x + tan x = sec x


    2. Relevant equations
    I can move terms around with identities, but I'm stuck with the partially solved equation below. I don't know how to solve a quadratic with weird terms. I got really far. But how do I show sin x?

    sin x = (-1 +- sqrt (1+9))/-4 = 1, -.5 which implies x = pi/2, 7pi/6, 11pi/6 ; range [0, 2pi).


    3. The attempt at a solution
    cos x (2 cos x + tan x) = sec x (cos x)
    2 cos^2 x + sin x = 1
    2 cos^2 x - 1 + sin x = 0
    (1 - 2 sin^2 x) + sin x = 0
    -2sin^2 x + sin x + 1 = 0
     
  2. jcsd
  3. Dec 7, 2008 #2

    rock.freak667

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    Homework Helper

    If you put t=sinx

    you will get -2t2+t+1=0

    Now use the quadratic equation formula
     
  4. Dec 7, 2008 #3
    Thanks Rock. I got confused with the solution given to me. It showed taking the root of 10 gives some kind of rational number. I was so lost. At least I know where I was messing up. Thanks.
     
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