General Solutions (1 Viewer)

[DJel]

Hi,

I am stuck on a couple of trigonometric general solution questions. They are;

• sin x + cos x = 1
• sqrt(3)cos x - sin x = 1

I am having some trouble as the answers do not seem to follow the general solutions at all, I can see that the answers work by drawing a graph, but I would like to know the non-graphical method.

Thanks,

DJel.

 

[cwag]

This is the way i do general solutions...

1) sin x + cos x =1
i change this into R sin (x + [FONT=&quot]α). where R=sqrt(a² + b[/FONT][FONT=&quot]²)
and [/FONT][FONT=&quot]α= tan[/FONT][FONT=&quot]^-1b/a
therefore we get: sqrt2 sin (x + [/FONT][FONT=&quot]п/4) =1
sin (x + [/FONT][FONT=&quot]п/4) = 1/sqrt2 (sin postive in 1st and 2nd quadrant)
x + [/FONT][FONT=&quot]п/4 = [/FONT][FONT=&quot]п/4 + 2k[/FONT][FONT=&quot]п[/FONT][FONT=&quot], 3[/FONT][FONT=&quot]п/4 + 2k[/FONT][FONT=&quot]п (where k= 0, [/FONT][FONT=&quot]±1, [/FONT][FONT=&quot]±2......)
therefore x= 2k[/FONT][FONT=&quot]п , [/FONT][FONT=&quot]п/2 + 2k[/FONT][FONT=&quot]п

2) sqrt3 cos x - sin x = 1
changes to 2 cos (x - [/FONT][FONT=&quot]п/6) = 1
cos (x-[/FONT][FONT=&quot] п/6) = 1/2 (cos postive in 1st and 4th quadrant)
x - [/FONT][FONT=&quot]п/6 = [/FONT][FONT=&quot]п/3 + 2k[/FONT][FONT=&quot]п, [/FONT][FONT=&quot]5п/3 + 2k[/FONT][FONT=&quot]п (where k = 0, [/FONT][FONT=&quot]±1, [/FONT][FONT=&quot]±2......)[/FONT]
[FONT=&quot] therfore x = [/FONT][FONT=&quot]п/6 + 2k[/FONT][FONT=&quot]п, 3[/FONT][FONT=&quot]п/2 + 2k[/FONT][FONT=&quot]п[/FONT][FONT=&quot]

Sorry if this is confusing. There may be an easier way, however this is my solution
[/FONT]

 

[namburger]

Transform them
1.
sin x + cos x = r Sin (x+a)
= r ( Sin x Cos a + Sin a Cos x)
Cos a = 1
Sin a = 1
Therefore Tan a = 1
a = pi/4
r = root ( 1 + 1)
= root 2
THEREFORE sin x + cos x = root 2 Sin (x+pi/4)

root 2 Sin (x+pi/4) = 1
Sin (x+pi/4) = 1/root 2
x+pi/4 = pi/4 + 2kpi OR x+pi/4 = 3pi/4 + 2kpi
Answer: x = 2kpi OR pi/2 + 2kpi (Where K is a constant)

2.
sqrt(3)cos x - sin x = r Cos (x+a)
= r (cos x cos a - sin x sin a)
Cos a = root 3
Sin a = 1
Therefore Tan a = 1/root 3
a = pi/6
r = root (3 + 1) = 2
THEREFORE sqrt(3)cos x - sin x = 2 Cos (x+pi/6)

2 Cos (x+pi/6) = 1
Cos (x+pi/6) = 1/2
x+pi/6 = pi/3 + 2kpi OR x+pi/6 = -pi/3 + 2kpi
Answer: x = pi/6 + 2kpi OR x = -pi/2 + 2kpi

 

[cwag]

well sure...jsut completly repeat me...

 

[undalay]

fuck up cwag. Namburger's final solution is slightly different from yours.
BOS is about helping people, sif try to hog the credit.

 

[DJel]

Sin (x+pi/4) = 1/root 2
x+pi/4 = pi/4 + 2kpi OR x+pi/4 = 3pi/4 + 2kpi

I have exactly the same working up until the bottom line in the quote. Why is it that we add 2kpi, instead of just finding the angles then putting them into the general solution (from textbook);

npi + (-1^n) x angle (where n is an integer)

?

 

[cwag]

haha...im sorry, thats true...i was just joking. and -pi/2 is same as [FONT=&quot]3[/FONT][FONT=&quot]pi/2.
[/FONT]

 

[undalay]

some people don't use the formula from the textbook.

edit: thats k cwag, and i see it is the same noww.

 

[DJel]

So I should just add 2kpi when I get a question requiring transformation (and finding a general solution) rather than the formulas?

Thanks for the help by the way guys,

DJel.

 

[cwag]

well, personally..the way we did it i find much easier. its essentially the same thing...we are adding 2kpi to the angle because k stands for the amount of times u go around in a 2pi revolution...what ever you think is the easier way. but personally, just finding the angles that work between 0 and 2pi, then adding 2kpi is easier.