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May 4, 2016
Posted by **Kate RESPONSE** on Monday, March 1, 2010 at 12:10am.

I don't follow your work Reiny

¡î3sinx + cosx = 1

¡î3cosx = 1-cosx

square both sides

3cos^2 x = 1 - 2cosx + cos^2 x

3(1-sin^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 = 1-cosx

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 12:24amHere'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 12:31amSorry 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 12:48amThanks 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 12:57amI 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 1:03amTHANKS