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Trig - General Solutions (help needed) (1 Viewer)

Petinga

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help with Trig equations - General Solutions

Find the general solutions to the following:

1. sin3x + sinx = 0

2. cos4x + cos2x = 0

3. sec2x = cosec2x

4. sin2x + sin4x = sin3x


i am in desperate need of help
 
Last edited:

Alexluby

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1. sin3x + sinx = 0
sin2x cosx + cos2x sinx +sinx = 0
2sinx cosx cosx + (cos^2 x - sin^2 x) sin x + sinx = 0
2cos^2 x + cos^2 x - sin^2 x +1 = 0
3cos^2x + cos^2x = 0
4cos^2x = 0

Ill do the questions one by one... but i'm not sure if im right either... just here to learn :p
 

shafqat

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Alexluby said:
1. sin3x + sinx = 0
sin2x cosx + cos2x sinx +sinx = 0
2sinx cosx cosx + (cos^2 x - sin^2 x) sin x + sinx = 0
2cos^2 x + cos^2 x - sin^2 x +1 = 0
3cos^2x + cos^2x = 0
4cos^2x = 0

Ill do the questions one by one... but i'm not sure if im right either... just here to learn :p
Looks right. It can also be done like this:
sin3x = -sinx
sin3x = sin-x
Then use general solutions for the sina = sinb case.
 

Alexluby

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Looks right. It can also be done like this: sin3x = -sinx sin3x = sin-x Then use general solutions for the sina = sinb case.
:S im lost with that.... do u think u cud explain it more?
 

SaHbEeWaH

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Alexluby said:
:S im lost with that.... do u think u cud explain it more?
something like..

sin3x = -sinx
sin3x = sin(-x)

if sinθ = sinα
then θ = nπ + (-1)nα

3x = nπ + (-1)n(-x)
3x - (-1)n(-x) = nπ
3x + (-1)nx = nπ
x(3 + (-1)n) = nπ
x = /(3 + (-1)n)

don't know if you can do that
i just made it up
 

Trebla

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2) cos4x + cos2x = 0

cos²2x - sin²2x + cos²x - sin²x = 0
cos²2x - 4sin²x.cos²x + cos²x - sin²x = 0
(2cos²x - 1)² - 4cos²x(1 - cos²x) + cos²x - (1 - cos²x) = 0
4cos^4 (x) - 4cos²x + 1 - 4cos²x + 4cos^4 (x) + cos²x - 1 + cos²x = 0
8cos^4 (x) - 6cos²x = 0
4cos^4 (x) - 3cos²x = 0
cos²x(4cos²x - 3) = 0
.: cos²x = 0 and cos²x = 3/4
.: x = 360n ± 90
AND x = 360n ± 30
Where n is an integer

3) sec2x = cosec2x

_1_ = _1_
cos2x sin2x

sin2x = cos2x
sin2x = 1
cos2x

.: tan 2x = 1
.: 2x = 180n + 45
.: x = 90n + 22.5
Where n is an integer

Please Check...
 
Last edited:

Slidey

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1. sin3x + sinx = 0
sin(a+b)=sinacosb+cosasinb
sin(a-b)=sinacosb-cosasinb
sina(a-b)+sin(a+b)=2sinacosb
a=2x
b=x
sin3x+sinx=2sin2xcosx=0
sin2x=0 or cosx=0
Go from there.

2. cos4x + cos2x = 0
cos(a+b)=cosacosb-sinasinb
cos(a-b)=cosacosb+sinasinb
cos(a-b)+cos(a+b)=2cosacosb
a=3x
b=x
cos4x+cos2x=2cos3xcosx=0
cos3x=0 or cosx=0
Ditto.

3. sec2x = cosec2x
Trebla did a solution to this.

4. sin2x + sin4x = sin3x
a=3x
b=x
2sin3xcosx=sin3x
sin3x(2cosx-1)=0
sin3x=0 or cosx=1/2
Go from there.
 

DRAGONZ

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Trebla said:
3) sec2x = cosec2x

_1_ = _1_
cos2x sin2x

sin2x = cos2x
sin2x = 1
cos2x

.: tan 2x = 1
.: x = 180n + 22.5
Where n is an integer
I debate this answer. Slide Rule can tell me if I'm right or not, but I think you've discounted half of the possible answers :confused: :confused:

Once you get to tan2x = 1:

2x = (pi/4) ± n(pi) --- n ξ J [pretend ξ = in the domain of)

.:. x = (pi/8) ± n(pi)/2 --- n ξ J

See, I get 90n, whereas you have 180n ...
 

FinalFantasy

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DRAGONZ said:
I debate this answer. Slide Rule can tell me if I'm right or not, but I think you've discounted half of the possible answers :confused: :confused:

Once you get to tan2x = 1:

2x = (pi/4) ± n(pi) --- n ξ J [pretend ξ = in the domain of)

.:. x = (pi/8) ± n(pi)/2 --- n ξ J

See, I get 90n, whereas you have 180n ...
hey i agree w\ u
tan2x=1
den 2x=kpi+pi\4
x=kpi\2+pi\8 where k is positive integer

by the way the person who wrote that is not Slide Rule
 

DRAGONZ

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FinalFantasy said:
hey i agree w\ u
tan2x=1
den 2x=kpi+pi\4
x=kpi\2+pi\8 where k is positive integer

by the way the person who wrote that is not Slide Rule
Hehe awesome. So we are right :D

And I know it wasn't Slide Rule. It was Treblar's solution. I merely asked Slide Rule to confirm that I was right ;)

But you've done the job for him.
 

FinalFantasy

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DRAGONZ said:
Hehe awesome. So we are right :D

And I know it wasn't Slide Rule. It was Treblar's solution. I merely asked Slide Rule to confirm that I was right ;)

But you've done the job for him.
oh i thought u thought that Slide Rule was the poster cuz u said
"...but I think you've discounted... " after mentioning Slide Rule
lol
Lets wait for Slide Rule to confirm both of our solutions before we conclude that we are right!:)
 

Trebla

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The person who asked this question by post is in YEAR 11. Your use of radians may cause some confusion.
Check my edit.
Once again, my working is criticised by a bunch of 4 unit Year 12s... :(
I'M ONLY IN YEAR 11!
 

FinalFantasy

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Trebla said:
The person who asked this question by post is in YEAR 11. Your use of radians may cause some confusion.
Check my edit.
Once again, my working is criticised by a bunch of 4 unit Year 12s... :(
I'M ONLY IN YEAR 11!
heyhey... we just posting to make sure it's right.. it isn't a criticism...
I'm sorry if it sounded like one..
 

Slidey

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DRAGONZ said:
Slide Rule can tell me if I'm right or not
I rather think I might easily have made the same mistake as Trebla. It's a simple transcription error.

But well spotted. :)
 

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