Posted by Kate RESPONSE on Monday, March 1, 2010 at 12:10am.
h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 6 7 4 1 7 1 8 8
I don't follow your work Reiny
¡î3sinx + cosx = 1
¡î3cosx = 1cosx
square both sides
3cos^2 x = 1  2cosx + cos^2 x
3(1sin^2 x) = 1  2cosx + cos^2 x
this reduces easily to
2cos^2 x  cosx  1 = 0
(2cosx + 1)(cosx  1) = 0
cosx = 1/2 or cosx = 1
x or theta = 120¨¬ , 240¨¬ or 0 , 360¨¬
x = 0, 2pi/3, 4pi/3, 2pi
Line 2 you subtracted cos x form both sides
¡î3sinx + cosx = 1
¡î3cosx = 1cosx
how did the sin x next to the SQRT(3) change to cos x magically...
I don't think I'm seeing something here that I should be...

RESPONSE to PreCalc Question  Kate RESPONSE, Monday, March 1, 2010 at 12:24am
Here's what I did
SQRT(3)sinx + cosx = 1
subtract cos x from both sides
SQRT(3)sinx = 1  cosx
squared both sides
3 sin^2 x = 1  2cosx +cos^2 x
replaced sin^2 x with a trig idendity
3 (1  cos^2 x) = 1  2cosx + cos^2 x
multiplied out the three
3  3cos^2 x = 1  2cosx + cos^2 x
subtracted cos^2 x from both sides
3  4cos^2 x = 1  2cosx
added 2cosx to both sides
3  4cos^2 x + 2 cosx = 1
subtracted 1 from both sides
2  4cos^2 x + 2cosx = 0
factored out the 2
2( 1  2cos^2 x + cos x) = 0
factored out the cosx
2(1 +cosx(2cosx + 1) = 0
2 = 0 or 1 +cosx(2cosx + 1)=0
from which I am stuck yet again... 
sorry Kate RESPONSE to PreCalc Question  Reiny, Monday, March 1, 2010 at 12:31am
Sorry Kate, I had written it out on paper and when I copied it to type I must have "jumped" lines.
Here is the good version:
√3sinx + cosx = 1
√3sinx = 1  cosx
square both sides
3sin^2 x = 1  2cosx + cos^2 x
3(1  cos^2 x) = 1  2cosx + cos^2 x
3  3cos^2 x = 1  2cosx + cos^2 x
4cos^2 x + 2cosx + 2 =0
2cos^2 x  cosx  1 = 0
Notice I had that, so the rest of my post is ok
I also included another reply to your first posting below suggesting the method your text is probably using. 
RESPONSE  RESPONSE, Monday, March 1, 2010 at 12:48am
Thanks i got to read that post...
but now that I got to this I don't see what is wrong with doing it this way but i guess it is important to solve both ways
heres what I did
4cos^2 x + 2cosx + 2 =0
factored
(4cosx  2)(cosx  1) = 0
4cosx  2 = 0 or cosx  1 = 0
4cosx  2 = 0
added 2 to both sides
4cosx = 2
divided by 4 on both sides
cosx = 1/2
this gives me two answers
(2 pi)/3, (4 pi)/3
cosx  1 = 0
added 1 to both sides
cosx = 0
x = 0
solutions
x = 0, (2 pi)/3, (4 pi)/3
now I checked my solutions with my magic number box (calculator) and oddly enough the third solution, (4 pi)/3, does not work but yet the other ones do...
I do not see why???
Well of to figuring out that other way...
I just don't understand what is wrong with the third solution and this way of solving is much easier I but I guess I got figure out the other way as well
THANKS 
RESPONSE to PreCalc Question  Reiny, Monday, March 1, 2010 at 12:57am
I explained why the second answer of 4pi/3 does not work in the reply to your first posting below
When you square both sides of an equation, extraneous roots are sometimes introduced. That is why all answers must be checked in the original equation, which is what I showed you and did.
finally your answers should be
0, 2pi/3, 2pi (0, 120 degrees, 360 degrees) 
RESPONSE to PreCalc Question  RESPONSE, Monday, March 1, 2010 at 1:03am
THANKS