and whyever is there a (-1)n in your:
x - \frac{\pi}{3}=n\pi + (-1)^{n}\frac{\pi}{6}
HallsofIvy
Sep7-08, 08:59 AM
1. The problem statement, all variables and given/known data
tan x + sec x=\sqrt{3}
Find x in 0 to 2*pi
3. 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]
so that
[tex] 4sin^2 x + 2sin x- 2=0
or
[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 x= \pi/6 or 5\pi/6 and if sin x= -1, then x= 3\pi/2.
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 [itex]\sqrt{3}! If [itex]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
Sep8-08, 10:08 AM
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]
so that
[tex] 4sin^2 x + 2sin x- 2=0
or
[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 x= \pi/6 or 5\pi/6 and if sin x= -1, then x= 3\pi/2.
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 [itex]\sqrt{3}! If [itex]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.
Thats obvious. But I used the polar format there.....
\sqrt{3}cos x - sin x=1
Suppose:
f(x)=a cos x+ b sin x
let
a =r sin y
b =r cos y
r=\sqrt{a^{2}+b^{2}}
y=tan^{-1}\frac{x}{y}
then f(x)=r sin(x+y)
I used this. And I am wondering what i did wrong????