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Solving trig equation for giving range

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data

    (b) Solve, for 0 <= x < 360°, the equation
    2tan²x + secx = 1,

    2. Relevant equations

    I think I have to solve it quadratically but I need to get the trigs to the same trig form

    3. The attempt at a solution

    [tex]2\frac{sin^{2}x}{cos^{2}x} + \frac{1}{cosx} = 1[/tex]

    but I cant see any other logical step after that?

    Thanks :)
  2. jcsd
  3. Sep 6, 2008 #2


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    How do you relate tan^2 x to sec^2 x ?
  4. Sep 6, 2008 #3


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    It would seem logical to me to multiply both sides by cos2 x: 2sin2 x+ cos x= cos2 x. Now replace sin2 x by 1- cos2 x and you have a quadratic equation for cos x.
  5. Sep 6, 2008 #4


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    Or, as Defennder was hinting :wink:, use one of the standard trigonometric identities :
    tan²x = sec²x - 1 :smile:
    Last edited: Sep 6, 2008
  6. Sep 6, 2008 #5
    yay ya i got it :) cheers :)
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