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Math Help - sec x + tan x = root 3

  1. #1
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    sec x + tan x = root 3

    Determine all solutions of the equation: sec x + tan x = root 3

    I put everything in terms of cos and sin and got:

    1/cosx + sinx/cosx = root 3

    1+sinx = root 3

    sinx = root 3 -1

    sinx = 0.73205

    x= 47.06

    Does anyone know if this is right so far? I'm not sure what to do next. The answers are in radians and laid out as follows: (k means all integers)

    x= pi/6+2pi(k)

    x= 5/6+2pi(k)

    x= 2pi(k)

    x= 7pi/6+2pi(k)

    x= 11pi/6 +2pi(k)


    Thanks so much for all help!
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  2. #2
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    Quote Originally Posted by daydreembelievr View Post
    Determine all solutions of the equation: sec x + tan x = root 3

    I put everything in terms of cos and sin and got:

    1/cosx + sinx/cosx = root 3

    1+sinx = root 3 what happened to cosx in the denominator?

    sinx = root 3 -1

    sinx = 0.73205

    x= 47.06
    \sec{x} + \tan{x} = \sqrt{3}

    \sec{x} = \sqrt{3} - \tan{x}

    square both sides ...

    \sec^2{x} = 3 - 2\sqrt{3} \tan{x} + \tan^2{x}

    1 + \tan^2{x} = 3 - 2\sqrt{3} \tan{x} + \tan^2{x}

    0 = 2 - 2\sqrt{3} \tan{x}

    \tan{x} = \frac{1}{\sqrt{3}}

    x = \frac{\pi}{6}

    x = \frac{7\pi}{6} is an extraneous solution
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