Question

One more peeps

Answer 1

\[\sin(2x) = sinx\]

Answer 2

Here's what I did:
2sinxcosx = sinx
2 sinxcosx - sinx = 0
sinx(2cosx - 1) = 0

Then I solved from there...
x = 0 + npi, x = pi/3 + 2npi and x = 2pi/3 + 2npi

Do those solutions look correct?

Answer 3

2pi/3 doesn't give positive 1/2.
I think it's a different angle for your third one.

Answer 4

Oh nuts! You're right.
5pi/3 + 2npi ?

Answer 5

Yayyy Pebs!

Answer 6

The 0 + npi looked good though?
Zebs :)

Answer 7

Also, should I be putting the plus or minus sign on these problems?

Answer 8

Yes the 0+npi looks good.
You can always double check your solutions using a graph.
I find it useful at least :)
https://www.desmos.com/calculator/txvvnr1h6z
The solutions are the intersections of these curves.

plus/minus?
It depends what you mean by n.

Answer 9

If you mean for it to be an integers, which is all positive and negative whole numbers, then no you're fine, \(\large\rm n=\{...,-3,-2,-1,0,1,2,3,...\}\)
\(\large\rm x=0+n\pi\)

But if you mean for n to be a natural number, then yes you need plus/minus to get those backwards spins, \(\large\rm n=\{0,1,2,3,...\}\)
\(\large\rm x=0\pm n\pi\)

Answer 10

That probably shouldn't say equals... blah bad maths

Answer 11

It really just depends what you've meant n to be this entire class :p

Answer 12

Hmmm. What if I don't specify what n is? What is it assumed to be?

Answer 13

I dunno :d

Answer 14

I dunno D:

Answer 15

"Here's what I did:
2sinxcosx = sinx
2 sinxcosx - sinx = 0
sinx(2cosx - 1) = 0"

That was excellently executed. Pebbleriffic job.

Answer 16

JINNNXXX

Answer 17

Lol, jinx

Answer 18

OMG

Answer 19

lol :3

Answer 20

haha xD

Answer 21

agent and zeppy are the kewlest helpers around here c:

Answer 22

Just say "where n is any integer"

Ty ty bebe ;)

Answer 23

I know :) pebbles is lucky

Answer 24

DON'T YOU SASS ME TSO!!!
DON'T MAKE ME BLOCK YOU AGAIN!!!

jk, i dunno why i did that the first time LOL

Answer 25

XD

Answer 26

btw you don't need to write

0 + n*pi

just n*pi

Answer 27

:O
really?

Answer 28

my whole life has been a lie

Answer 29

lol :D

Answer 30

ZEP!!! Why did you tell me?

Answer 31

*blushy face*

Answer 32

My bad D: I guess I thought that was obvious, ugh >.<

Answer 33

Cuz like.... from kindergarden.. i remember that \(0+4\)
is like... the same as ... \(4\)

Answer 34

Well 0 + pi is still pi, so...

Did @zepdrix really block my bebe @TheSmartOne at some point?!

Answer 35

LOL "from kindergarten" :D

Answer 36

:D well..

Answer 37

When I'm typing typing typing a response,
and someone sneaks in and just ... hands them the answer...
ya that rustles my jimmies real good :P lol
i get grumpy sometimes XD haha

Answer 38

Grampa Zep can get real grumpy if you tick him off.

Answer 39

:P

I actually majored in Jim-Rustlery in college,
you would think i could manage my jimmies pretty well...

I guess that money went to waste.

Answer 40

ahahah!

Answer 41

sin(2x) = sin(x)

2x = x + 2npi

can't we say that directly ?

Answer 42

"majored in Jim-Rustlery " :D

@ganeshie8 doesn't that miss solutions? And i'm trying to follow your logic of how you say that...

Answer 43

Alrighty guys, I'm retiring...
and by retiring, I mean hitting the sack...
... going to bed.
I dunno how you zombies stay up 24/7. But thank you so much for your help Zeb and Angel. <3

Answer 44

Feel free to continue the chatter though :D

Answer 45

Yeah, I think we can get the remaining solutions by using sin(x) = sin(pi-x)

Answer 46

sin(2x) = sin(x)

2x = x + 2npi OR 2x = pi-x + 2npi

Answer 47

that should get us all the solutions..

Answer 48

oo clever

Answer 49

Welcome Pebbles <3

sin(2x) = sin(x)

2x = x + 2npi

I kinda see how you got this jump. Since when x= n*pi, the equation is true... but it just seems like it takes more thought and effort to do it this way.

Answer 50

But...

2x = pi-x + 2npi

3x = pi + 2npi

x = pi/3 + 2npi/3

Answer 51

Yeah, but I think its worth spending the time and effort as it gives more insight into how periodic function work and to easily approach other messy equations like :

sin(5x) = sin(x)

Answer 52

I mean time and effort of the OP, not us haha

Answer 53

Ah, true.

Haha yes.

But these don't work here, do they?

2x = x + 2npi OR 2x = pi-x + 2npi

Answer 54

those are all the solutions, right ?

Answer 55

2x = x + 2npi OR 2x = pi-x + 2npi

after rearranging gives

x = 2npi OR x = pi/3 + 2npi/3

Answer 56

I *think* it was this: i corrected abbles mistake

x = npi, x = pi/3 + 2npi and x = 5pi/3 + 2npi

Answer 57

Ohk, I see... but they both must be same

Answer 58

I mean, your expression and my expression must yield the same solution set.

Answer 59

Ah yes.. maybe it does actually,

x = 2npi OR x = pi/3 + 2npi/3

x= 0, pi/3, pi, 5pi/3, 2pi, etc.

Answer 60

Nice :)

Answer 61

similar discussion here
http://math.stackexchange.com/questions/159244/what-is-the-correct-way-to-solve-sin2x-sinx

Answer 62

Interesting. Stackexchange often has impressive display of knowledge.

Answer 63

And as they point out, it would nicely work for any equation of the form \( \large\sin(a x)= \sin(b x)\)

Answer 64

\[\large\sin(a x)= \sin(b x) \]would have solutions

ax = bx + 2npi OR ax = pi - bx + 2npi

right?

Answer 65

I think so..
I don't think they can have any other solutions...

\[\sin(\clubsuit) = \sin( \color{red}{\spadesuit} )\\~\\ \implies \clubsuit = \color{red}{\spadesuit}+ 2n\pi ~~\lor~~ \clubsuit = \pi-\color{red}{\spadesuit}+ 2n\pi \]

Answer 66

Yeah cos it has to be when ♣=♠, and then you can just add the periodic factor, and use the pi-♠ identity.

Should work the same way for cosine and tangent too.

Answer 67

Why did you make the spade red... lol, are you not a poker or card player @ganeshie8? :P

Answer 68

Just checking which one of us was alert XD

Answer 69

lol XD it looked like a heart till i copied ut