Stupid Q - general solutions (1 Viewer)

[karnage]

Hi guys, having one of those stupid moments.

find the general solution of:

cos x = - 1/2

so i do

x = pi - pi/3, pi + pi/3, 3pi - pi/3, 3pi + pi/3...........................

I list it that way but i cant seem to get the general solution that the answer lists from analysing the pattern above. Heck, is that even the right list of values.

I know theres a formula which gives you the answer straight away, but i prefer to learn how to do it manually

 

[conics2008]

same i prefer manually, you need to imagine, first of all its -1/2 that means quadrant 2 and 3

find the angles there when you do those are the solutions for the 1st revolution.

if you want to go for another solution do this 360+ one of the angel and for the 2nd angel 360+ angel.. get the picture..

that means every revolution you will have 2 angel....

i dont have a calculator soo i cant be stuffed..

its the same aas general formula it use n for number of revolution.

 

[conics2008]

Ok, Ive got a calculator.

cos=-1/2 = 120 ( 2nd Quadrant ) >> 240 ( 3rd Quadrant )

This is for the first revolution that is one round.

if you want to go for another round all you do is 360 + 120 and 360+240 and so on..

3rd round is 720 + 120 ... and soo on....

make life easier right.. see the simple logic in this

 

[ronnknee]

Okay, let's do this one step at a time

We look at this:
x = pi - pi/3, pi + pi/3, 3pi - pi/3, 3pi + pi/3, 5pi - pi/3, 5pi + pi/3 ...

And we notice the alternating plus and minus signs
Therefore we can simplify this to

x = pi ± pi/3, 3pi ± pi/3, 5pi ± pi/3 ...

Now look at the coefficients of pi
Notice how they are in an arithmetic progression?
ie. 1, 3, 5 ... with a difference of 2
Therefore, we can simplify that to 1 + 2n where n = 0, 1, 2, 3 ...

x = (1 + 2n)pi ± pi/3

That's the general solution
I hope you understand me

 

[mick135]

Okay, let's do this one step at a time

We look at this:
x = pi - pi/3, pi + pi/3, 3pi - pi/3, 3pi + pi/3, 5pi - pi/3, 5pi + pi/3 ...

And we notice the alternating plus and minus signs
Therefore we can simplify this to

x = pi ± pi/3, 3pi ± pi/3, 5pi ± pi/3 ...

Now look at the coefficients of pi
Notice how they are in an arithmetic progression?
ie. 1, 3, 5 ... with a difference of 2
Therefore, we can simplify that to 1 + 2n where n = 0, 1, 2, 3 ...

x = (1 + 2n)pi ± pi/3

That's the general solution
I hope you understand me

bingo - right on the money
you even nailed it when n = 0 - well done

 

[tommykins]

It's like the 2kpi method when doing roots of unity in complex numbers.

 

[karnage]

Gotcha, didnt think of using the arithmetic progressions in this. Cheers

 

[undalay]

In the exam, they won't penalise you if you have more then one general solution formula.

You can have 1 or 2 or 3 overlapping eachother, as long as they are all viable.

 

[namburger]

General solutions:
Sin
Sin x = @
x = arcsin @ + 2npi
AND x = (pi - arcsin @) + 2npi

Cos
Cos x = @
x = +/- arccos @ + 2npi

Tan
Tan x = @
x = arctan @ + npi

Where arc = inverse and n = any integer. sub it into a calculator and there are your general solutions

 

[qqmore]

General solutions:
Sin
Sin x = @
x = arcsin @ + npi
AND x = (pi - arcsin @) + 2npi

isnt it 2npi

 

[namburger]

isnt it 2npi

yep ill fix, sorry, i missed it

 

[karnage]

Yup those are the formulas i was talking about which gives you the answer straight away

Much easier i know, but i sorta want to learn it, then maybe start using the formulas

 

[undalay]

Finding patterns in some sequence isn't really learning how the general solutions formulas are constructed.

 

[lacklustre]

This is an incredibly stupid question from me (which didn't warrant the creation of a new thread just for it - so i'll stick it here). Please help me solve this simple inequality:

x²-1 > 0 (> = greater than or equal to in this case).

What i do:
x² > 1
x > plus/minus 1

therefore x > 1 , and x > -1

I don't know why i am incapable of getting the right answer (having a brain freeze) please don't laugh at me too hard .

 

[lsam]

This is an incredibly stupid question from me (which didn't warrant the creation of a new thread just for it - so i'll stick it here). Please help me solve this simple inequality:

x²-1 > 0 (> = greater than or equal to in this case).

What i do:
x² > 1
x > plus/minus 1

therefore x > 1 , and x > -1

I don't know why i am incapable of getting the right answer (having a brain freeze) please don't laugh at me too hard .

x^2 - 1 > 0
(x+1)(x-1) > 0
(Here, draw the graph)
Threfore, x> 1 or x<-1

 

[lyounamu]

x^2 - 1 > 0
(x+1)(x-1) > 0
(Here, draw the graph)
Threfore, x> 1 or x<-1

It should be >= in your case.

 

[lacklustre]

x^2 - 1 > 0
(x+1)(x-1) > 0
(Here, draw the graph)
Threfore, x> 1 or x<-1

i still dunno how you got x < -1

So is it absolutely necessary to draw the graph?

 

[lyounamu]

i still dunno how you got x < -1

So is it absolutely necessary to draw the graph?

He got x<-1 by graph-sketching.

By the way, it is not necessary to use graph. However, you need to test different points to make sure you are at the right domain.

 

[lacklustre]

He got x<-1 by graph-sketching.

By the way, it is not necessary to use graph. However, you need to test different points to make sure you are at the right domain.

Ah ok, this stuff is slowly coming back to me (it's been a while since i've worked with inequalities).

So if you get a harder inequality (that's harder to sketch), you would just test points then?

 

[lyounamu]

Ah ok, this stuff is slowly coming back to me (it's been a while since i've worked with inequalities).

So if you get a harder inequality (that's harder to sketch), you would just test points then?

Yeah. Testing is so much easier than drawing graph when it comes to complex inequality.