# sec x + tan x = root 3

• Dec 7th 2008, 11:58 AM
daydreembelievr
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! :D
• Dec 7th 2008, 02:07 PM
skeeter
Quote:

Originally Posted by daydreembelievr
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

$\displaystyle \sec{x} + \tan{x} = \sqrt{3}$

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

square both sides ...

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

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

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

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

$\displaystyle x = \frac{\pi}{6}$

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