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Tan(x) + sec(x) = sqrt(3), find x

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]tan x + sec x=\sqrt{3}[/tex]
    Find x in 0 to 2*pi

    3. The attempt at a solution
    [tex]\frac{sin x+1}{cos x}=\sqrt{3}[/tex]

    [tex]\sqrt{3}cos x - sin x=1[/tex]

    [tex]2 sin (x - \frac{\pi}{3})=1[/tex]

    [tex]x - \frac{\pi}{3}=n\pi + (-1)^{n}\frac{\pi}{6}[/tex]

    My problem is that the the solution x=pi/6 is missing fom my general solution. Why?????
  2. jcsd
  3. Sep 7, 2008 #2


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    Hi ritwik06! :smile:
    Nooo … [tex]\ \ 2\,sin (\frac{\pi}{3}\ -\ x)\ =\ 1[/tex] :smile:

    and whyever is there a (-1)n in your:
  4. Sep 7, 2008 #3


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    Re: Trigonometry

    It's hard to tell why if you don't show all of your work!

    How did you get from
    [tex]\sqrt{3}cos x - sin x=1[/tex]
    [tex]2 sin (x - \frac{\pi}{3})=1[/tex]?
    I would have done this a completely different way:
    [tex]\frac{sin x+1}{cos x}=\sqrt{3}[/tex]
    [tex]sin x+ 1= \sqrt{3} cos x[/tex]
    Now square both sides:
    [tex](sin x+ 1)^2= sin^2 x+ 2sin x+ 1= 3 cos^2x= 3(1- sin^2 x)[/itex]
    so that
    [tex] 4sin^2 x + 2sin x- 2=0[/tex]
    [tex]2 sin^2 x+ sin x- 2= (2 sin x- 1)(sin x+ 1)= 0.
    That has the two roots sin x= 1/2 and sin x= -1.

    If sin x= 1/2, then [itex]x= \pi/6[/itex] or [itex]5\pi/6[/itex] and if sin x= -1, then [itex]x= 3\pi/2[/itex].

    Since we squared, we may have introduced a new solution so we had better check in the original equation. If [itex]x= \pi/6[/itex], then [itex]tan x= sin x/cos x= \sqrt{3}/3[/itex] and [itex]sec x= 1/cos x= 2\sqrt{3}/3. Yes, those add to [itex]\sqrt{3}! If [itex]x= 5\pi/6[/itex] then [itex]tan x= sin x/cos x= -\sqrt{3}{3}[/itex], [itex]sec x= 1/cos x= -2\sqrt{3}/3[/itex] and those add to [itex]-\sqrt{3}[/itex], not [itex]\sqrt{3}[/itex]. If [itex]x= 3\pi/2[/itex], cos x does not exist and neither tan x nor sec x exists. The ONLY solution to the equation between 0 and [itex]2\pi[/itex] is [itex]x= \pi/6[/itex].
  5. Sep 8, 2008 #4
    Re: Trigonometry

    Thats obvious. But I used the polar format there.....
    \sqrt{3}cos x - sin x=1[/tex]

    f(x)=a cos x+ b sin x
    a =r sin y
    b =r cos y

    then f(x)=r sin(x+y)
    I used this. And I am wondering what i did wrong????
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