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

[ritwik06]

Homework Statement

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

The Attempt at a Solution

\frac{sin x+1}{cos x}=\sqrt{3}


\sqrt{3}cos x - sin x=1


2 sin (x - \frac{\pi}{3})=1


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

My problem is that the the solution x=pi/6 is missing fom my general solution. Why?

 

[tiny-tim]

Hi ritwik06!

\sqrt{3}cos x - sin x=1

2 sin (x - \frac{\pi}{3})=1

Nooo … \ \ 2\,sin (\frac{\pi}{3}\ -\ x)\ =\ 1

and whyever is there a (-1⁾ⁿ in your:

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

 

[HallsofIvy]

Homework Statement

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

The Attempt at a Solution

\frac{sin x+1}{cos x}=\sqrt{3}


\sqrt{3}cos x - sin x=1


2 sin (x - \frac{\pi}{3})=1


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

My problem is that the the solution x=pi/6 is missing fom my general solution. Why?

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

How did you get from
\sqrt{3}cos x - sin x=1
to
2 sin (x - \frac{\pi}{3})=1?
I would have done this a completely different way:
From
\frac{sin x+1}{cos x}=\sqrt{3}
sin x+ 1= \sqrt{3} cos x
Now square both sides:
(sin x+ 1)^2= sin^2 x+ 2sin x+ 1= 3 cos^2x= 3(1- sin^2 x)[/itex]<br /> so that<br /> 4sin^2 x + 2sin x- 2=0<br /> or<br /> 2 sin^2 x+ sin x- 2= (2 sin x- 1)(sin x+ 1)= 0.&lt;br /&gt; That has the two roots sin x= 1/2 and sin x= -1.&lt;br /&gt; &lt;br /&gt; If sin x= 1/2, then x= \pi/6 or 5\pi/6 and if sin x= -1, then x= 3\pi/2.&lt;br /&gt; &lt;br /&gt; Since we squared, we may have introduced a new solution so we had better check in the original equation. If x= \pi/6, then tan x= sin x/cos x= \sqrt{3}/3 and sec x= 1/cos x= 2\sqrt{3}/3. Yes, those add to \sqrt{3}! If x= 5\pi/6 then tan x= sin x/cos x= -\sqrt{3}{3}, sec x= 1/cos x= -2\sqrt{3}/3 and those add to -\sqrt{3}, not \sqrt{3}. If x= 3\pi/2, cos x does not exist and neither tan x nor sec x exists. The ONLY solution to the equation between 0 and 2\pi is x= \pi/6.

 

[ritwik06]

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

How did you get from
\sqrt{3}cos x - sin x=1
to
2 sin (x - \frac{\pi}{3})=1?
I would have done this a completely different way:
From
\frac{sin x+1}{cos x}=\sqrt{3}
sin x+ 1= \sqrt{3} cos x
Now square both sides:
(sin x+ 1)^2= sin^2 x+ 2sin x+ 1= 3 cos^2x= 3(1- sin^2 x)[/itex]<br /> so that<br /> 4sin^2 x + 2sin x- 2=0<br /> or<br /> 2 sin^2 x+ sin x- 2= (2 sin x- 1)(sin x+ 1)= 0.&lt;br /&gt; That has the two roots sin x= 1/2 and sin x= -1.&lt;br /&gt; &lt;br /&gt; If sin x= 1/2, then x= \pi/6 or 5\pi/6 and if sin x= -1, then x= 3\pi/2.&lt;br /&gt; &lt;br /&gt; Since we squared, we may have introduced a new solution so we had better check in the original equation. If x= \pi/6, then tan x= sin x/cos x= \sqrt{3}/3 and sec x= 1/cos x= 2\sqrt{3}/3. Yes, those add to \sqrt{3}! If x= 5\pi/6 then tan x= sin x/cos x= -\sqrt{3}{3}, sec x= 1/cos x= -2\sqrt{3}/3 and those add to -\sqrt{3}, not \sqrt{3}. If x= 3\pi/2, cos x does not exist and neither tan x nor sec x exists. The ONLY solution to the equation between 0 and 2\pi is x= \pi/6.

&amp;amp;lt;br /&amp;amp;gt; Thats obvious. But I used the polar format there...&amp;amp;lt;br /&amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; \sqrt{3}cos x - sin x=1&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Suppose:&amp;amp;lt;br /&amp;amp;gt; f(x)=a cos x+ b sin x&amp;amp;lt;br /&amp;amp;gt; let &amp;amp;lt;br /&amp;amp;gt; a =r sin y &amp;amp;lt;br /&amp;amp;gt; b =r cos y&amp;amp;lt;br /&amp;amp;gt; r=\sqrt{a^{2}+b^{2}} &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; y=tan^{-1}\frac{x}{y}&amp;amp;lt;br /&amp;amp;gt; then f(x)=r sin(x+y) &amp;amp;lt;br /&amp;amp;gt; I used this. And I am wondering what i did wrong?