Find all solutions in the interval [0, 2π).
sin2 x + sin x = 0

- anonymous

Find all solutions in the interval [0, 2π).
sin2 x + sin x = 0

- Stacey Warren - Expert brainly.com

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- chestercat

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- Psymon

We cannot have different angles when we solve for these. We have a 2x and an x, and we have to have them be the same. So what we want to do is use an identity on sin2x so we can get functions with only an angle of x. Do you know the double-angle formula for sin2x?

- Luigi0210

Is that:
\[\sin^2x\]
Or \[\sin2x\]

- Psymon

Yeah, completely different if thats supposed to be squared xD Which I kinda think it is.

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## More answers

- anonymous

yes, it is squared.

- Luigi0210

If it's squared then it's simple simple simple

- Psymon

Alrighty. Then this we would factor in the same way we would as if those sin(x)'s were just regular x's. We factor it just like we woudl this:
\[x ^{2}+x=0 \]

- Psymon

So if you know how to factor the above quadratic, then you can factor this sin^2x + sinx = 0 equation :3

- anonymous

2 sin(x/2) cos(x/2) (sin(x/2)+cos(x/2))^2 = 0

- Psymon

Why did we do that? All we need to do is factor it like a quadratic. Its done in the same fashion as something like this:
\[x ^{3}+x=x(x ^{2}+1)\]
Youd be doing that process with your equation :3

- anonymous

Oh. I tend to over think things...

- Psymon

Well, youre just not used to seeing a quadratic being with trig functions xD

- anonymous

Exactly.

- Psymon

Thats why trig is tricky. Nothing like all the tons of algebra we have to do to get here :p SO following the method I did above, think you can factor what we have with our sines?

- AkashdeepDeb

Is this query cleared?

- Psymon

We're still working it xD

- AkashdeepDeb

Any 'help' needed?

- Psymon

I dont believe so. Just waiting for her to respond to see where we might need to go from here.

- AkashdeepDeb

xD

- anonymous

I'm lost, I don't know what factor.

- anonymous

to factor

- Psymon

Ah, okay. Well, akash is eager, lol.

- anonymous

Hopefully, Akash can make this problem a bit less complicated like the previous one.

- Psymon

Yeah, you have a hard time understanding me it seems. Im sure hell do much better than what I can do.

- AkashdeepDeb

1 sec.

- anonymous

No, it's just that these problems are really difficult for me. Don't take it personally.

- AkashdeepDeb

Guys chill out on that!! LOL

- Psymon

No, its not that I take it persoanlly, I just think I suck compared to everyone else. So usually once someone else joins in ont he problem I leave because Im just going to embarrass myself.

- AkashdeepDeb

All you have to do is take
sin x(sin x + 1) = 0
Now either
sin x = 0
Or sin x = -1
So x can be
0' or 270'

- anonymous

Yeah, I really despise any type of conflict...

- AkashdeepDeb

'LoveIsKey' after all

- Psymon

Which is why its usually best for me to leave when others come in and join. I tend to cause a lot of conflict and drama.

- anonymous

@Psymon You should not think like that. This is not the first time you have aided me in a math problem, you're amazing. Do not underestimate yourself !

- AkashdeepDeb

PS. Psymon you have a score of 74 now!
PROOF THAT YOU DON'T SUCK!! :D

- Psymon

that means nothing, lol. It means I go to a lot of problems xD

- AkashdeepDeb

LoL, No chill out on the 'Insulting yourself' XD :D
It is FUNNY

- AkashdeepDeb

Okay, Not drifting away from the topic
@Loveiskey18
Did you understand what I did?

- Psymon

Lol, it what I always do xD

- anonymous

How did you get 270?

- AkashdeepDeb

sin 270 is -1
That's how! :)

- anonymous

I'm just going to agree with you... -__-

- Psymon

Sin refers to y-coordinates. If you actually look at the circle and point our where 270 is....
|dw:1377670372265:dw|

- AkashdeepDeb

Why why why???
Ask me if you have any question!!!!

- AkashdeepDeb

Yeah! Psymon is right over there!

- anonymous

This sin, cos, tan stuff is the reason why I can not get an amazing score on my ACT . Why do we have them, because I am really going to use sin, cos and tan in my daily life.. :)

- Psymon

Well, a lot of people think math is a sin.

- AkashdeepDeb

Try Learning all
cosine, sine and tangent values for
0, pi/2 , pi , 3pi/2 , 2pi radians!
It WILL help!! :D

- Psymon

And the people who love math say without geometry life is pointless.

- anonymous

Oh, the unit of Circle !

- AkashdeepDeb

Lol There!!

- AkashdeepDeb

@Loveiskey18 Did you get how I computed this though?

- Psymon

Yep yep. Absolutely. All the values on the unit circle refer to actual points on a graph. Sin(pi/6) is 1/2 because the y-coordinate, if you were to plot it, is 1/2!

- anonymous

Yes, somewhat. Laughing out loud

- AkashdeepDeb

What is the other 'Somewhat' you do not get?

- anonymous

Oh, I'm just wondering how we are going to get from 270 to this:
x = 0, π, pi divided by three, five pi divided by three
x = 0, π, pi divided by three, two pi divided by three
x = 0, π, four pi divided by three, five pi divided by three
x = 0, π, three pi divided by two.
That is the somewhat that I do not get.

- AkashdeepDeb

Btw I forgot to mention even
180 degrees would be a value because
sin 180 or sin pi is 0 :D
So it should be Option D.!!! :D

- Psymon

Makes me think of the little cartoons in my math book:
|dw:1377670698620:dw|

- AkashdeepDeb

@Psymon LOLLLLL
@Loveiskey18
Get it?
Just learn some trig values! :D

- anonymous

Yes, actually I do, Thank You. Do you mind helping me with 3 more problems, I've been at these 35 questions for 5 hours and counting.
@Psymon What is the matter with you? Laughing out loud !!

- Psymon

Also, when you see pi in your answers, you are usually dealing with radian measures. We know we had an answer that was 270, but we had radian answers. If you dont know what a degree measure is in radians, you can do multiply:
\[degrees*\frac{ \pi }{ 180 } \]
\[270*\frac{ \pi }{ 180 }=\frac{ 3\pi }{ 2 }\]

- AkashdeepDeb

First, You have to try it yourself and ONLY THEN if you don't get it will I HELP YOU! :D

- anonymous

Don't you think I tried, I did mention I've been working on these problems for 5 hours, and that is not an exaggeration.

- AkashdeepDeb

Cool! Ask away! :D

- anonymous

I hope you can handle these problems...

- AkashdeepDeb

So do I. :D

- anonymous

Verify the identity.
sin (x + y) - sin (x - y) = 2 cos x sin y

- AkashdeepDeb

Expand this using the identities!
sin(x+y) = sinx.cosy + siny.cosx
sin(x-y) = sinx.cosy - siny.cosx
Now subtracting them we get
sinx.cosy + siny.cosx - sinx.cosy + siny.cosx
2 cosx.siny
Understood? :)

- Psymon

Identities take practice to be fair xD We probably can all do them, but we also probably sucked at them when we first did it xD

- anonymous

I did this:
sin(x+y)-sin(x-y) = 2 cos(x) sin(y)
cos(x) sin(y) = 1/2 (sin(x+y)-sin(x-y)) = 1/2 (sin(x+y)-sin(x-y))

- anonymous

I know I suck at them . . .

- AkashdeepDeb

A wee-bit complicated don't you think!
Take A break!
You've been at it for 5 hours!
That's how much I study altogether in 1 WEEK!

- Psymon

A general kind of rulethey try and have you do is to only mess with one sideof the equation. Pick the part that looks the most complicated and play around with thatside. Only mess with the other side if you get stuck. Because otherwise you do random identities with no rhyme and reason xD

- anonymous

sin(x+y)-sin(x-y) = 2 cos(x) sin(y)
cos(x) sin(y) = 1/2 (sin(x+y)-sin(x-y)) = 1/2 (sin(x+y)-sin(x-y))
sin(x+y)-sin(x-y) = ^sin(x+y)-sin(x-y)
The left and right hand match !

- Psymon

Right. But you dont actually prove the identity until you get one of the sides to match what you started with. You cant get them to equal at an intermediate step.

- anonymous

I can't I have postponed this for a week now, and I have to do it to move on. I am hoping to get atleast a 95 on this.

- Psymon

Getting themto equal in between is what helps you finally get there maybe, though :P

- AkashdeepDeb

And I believe you WILL!
Any other question?

- anonymous

@Psymon Wait, what are you talking about? I started with the right side then got what was on the left.

- Psymon

Oh, sorry xD I saw a bunch of random stuff and didnt realize what was what xD

- anonymous

Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the function f).
k = 4; f(x) = 4x4 - 3x3 - 2x2 - 5x - 4; Upper bound?
Does this question have something to do with the reminder?

- anonymous

It's okay :)

- AkashdeepDeb

@Loveiskey18
You wrote this right!
cos(x) sin(y) = 1/2 (sin(x+y)-sin(x-y)) = 1/2 (sin(x+y)-sin(x-y))
And said that they are equal is it?

- anonymous

Just let it be, I desire to be done.

- AkashdeepDeb

Just for info: That is not all totally correct btw!

- AkashdeepDeb

Where is k in your second question?

- anonymous

Wait, what do you mean.

- anonymous

and why is it not correct....?

- Psymon

Wow, not used to seeing bounds with those questions O.o

- AkashdeepDeb

Why do you think it IS CORRECT?
@Loveiskey18

- AkashdeepDeb

The question is to verify them!
You can write it like this!
cos(x) sin(y) = 1/2 (sin(x+y)-sin(x-y)) = 1/2 (sin(x+y)-sin(x-y))
But what you did was just right it again here ^^^^ !
So you cannot really right it again and prove that LHS=RHS!

- AkashdeepDeb

*write

- anonymous

Okay, wrote it just as you did.

- AkashdeepDeb

But the main thing is to understand it!
I hope you understood it and did it.

- Psymon

I can understand working on something all day and at the end not caring if I get it, lol. But yeah, best you understand it if possible xD

- AkashdeepDeb

@Psymon help her with the second one!
@Loveiskey18 Can you post the Last question?

- anonymous

I agree with Psymon, because that is exactly how I FEEL. Laughing out loud.

- Psymon

Whats the 2nd one? xD

- AkashdeepDeb

- Psymon

I can do the whole synthetic division, but Ive never even seen upper and lower bound be used in these questions O.o

- AkashdeepDeb

@Loveiskey18 POST THE NEXT QUESTION!

- anonymous

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
f(x) = quantity x minus nine divided by quantity x plus five. and g(x) = quantity negative seven x minus eight divided by quantity x minus one.|dw:1377672943415:dw|

- AkashdeepDeb

Your f(x) does not match the text!
Which one do I follow?

- anonymous

Disreard the thing that is written out. The question and drawing are correct. That is what happens when I am rushed...

- anonymous

Disregard

- AkashdeepDeb

Alright so following the drawing!
The question gives us a big clue!
We have to prove: that f(g(x)) = x and g(f(x)) = x.
Now do this....
|dw:1377673320194:dw|

- anonymous

Don't hate me if I say no, my apologies, but I am halfway asleep. :)

- AkashdeepDeb

What do you not get here?
Sleep now, Try it out tomorrow~'It is MORNING here!!'

- anonymous

I cannot exit out of it, so it's either I submit it now or continue working. It's 3:09 here.

- AkashdeepDeb

OMG!!
See here all they are asking us to do is: f(g(x)) = g(f(x)).
Just substitute the values of the functions values in the x's of the OTHER FUNCTIONS!!
Do it tomorrow, if you do not get it now.

- AkashdeepDeb

By OMG I meant 'Holy, it is 3:09 and you're still awake' :)

- Psymon

Okay, lets just say we had this
\[f(x) = x ^{2}+2x-5\]If I then asked you to give me f(2), would you know what to do ?

- anonymous

Hold on...

- Psymon

mkay

- AkashdeepDeb

Bye! I hope my answers helped!

- Psymon

later :3

- anonymous

Thank You !

- Psymon

Did it kind of make sense what I was asking?

- AkashdeepDeb

:)

- anonymous

Yes, g(x-8)/(x+7)=x, f(-7x - 8)/(x - 1) =x

- anonymous

f(x) = x-8/x+7
f((-7x-8/x-1) = [ (-7x-8)/(x-1) - 8 ] / [ (-7x-8)/(x-1) + 7 ] = -15x/-15 = x

- Psymon

Now you have to do it the other way and show that :3 g(f(x))

- anonymous

For some odd reason, I can not get g(f(x))

- Psymon

Well, its always possible they are not inverses. But lets see what I get : )

- anonymous

Okay.

- anonymous

This is what I have so far for g(f(x))=
g(x-8)/(x+7)

- Psymon

\[\frac{ -7(\frac{ x-8 }{ x+7 })-8 }{ \frac{ x-8 }{ x+7 }-1 }\]
\[\frac{ \frac{ -7(x-8)-8(x+7) }{ x+7 } }{ \frac{ x-8-(x+7) }{ x+7 } }\]
\[\frac{ -7(x-8)-8(x+7) }{ x-8-(x+7) }\]
\[\frac{ -7x+56-8x-56 }{ x-8-x-7 }=\frac{ -15x }{ -15 }=x \]

- anonymous

I don't think I would have gotten as far as you did, however I was on the right tract. Thank You !!!! By the way I did solve that upper bound question. Thank You once again, a thousand times to you and Akash. You guys are amazing :D

- Psymon

Lol, thats good then xD Hopefully you see what I did for it :P

- anonymous

I do, Thank You. Now I can finally SLEEP ! :)

- Psymon

Lol, sleep well then :3

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