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:
[tex](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 /> [tex]2 sin^2 x+ sin x- 2= (2 sin x- 1)(sin x+ 1)= 0.<br /> That has the two roots sin x= 1/2 and sin x= -1.<br /> <br /> If sin x= 1/2, then $x= \pi/6$ or $5\pi/6$ and if sin x= -1, then $x= 3\pi/2$.<br /> <br /> 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 [itex]sec x= 1/cos x= 2\sqrt{3}/3. Yes, those add to [itex]\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 $x= \pi/6$.[/itex][/itex][/tex][/tex]

 

[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:
[tex](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 /> [tex]2 sin^2 x+ sin x- 2= (2 sin x- 1)(sin x+ 1)= 0.<br /> That has the two roots sin x= 1/2 and sin x= -1.<br /> <br /> If sin x= 1/2, then $x= \pi/6$ or $5\pi/6$ and if sin x= -1, then $x= 3\pi/2$.<br /> <br /> 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 [itex]sec x= 1/cos x= 2\sqrt{3}/3. Yes, those add to [itex]\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 $x= \pi/6$.[/itex][/itex][/tex][/tex]

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