Question

sin 2x + sin x = 0
Find the exact solutions of the equation in the interval [0, 2π)
Dont really understand this, anyone have any answers?

Answer 1

\[\sin2x+\sin x=0\]
or
\[\sin^2x+\sin x=0\]

Answer 2

is it \(\sin^2(x)\) or \(\sin(2x)\) ?

Answer 3

lol

Answer 4

Gotta be sure :)

Answer 5

This is a practice one and the answer
sin 2x − sin x = 0

0, pie/3, pie, 5pie/3

I dont get how they got it

Answer 6

That's not what I asked... Is the sine squared, or is the x multiplied by 2?

Answer 7

mmmm pie \(\pi\)

Answer 8

your top answer

its sin 2x + sin x =0

Answer 9

It can't be sin squared; the answers to that are pi, 2pi, and pi/2

Answer 10

It is given to you that : \(\sin(2x) + \sin x = 0\)
We can write that as : \(\sin(2x) = -\sin (x)\)

Right?

Answer 11

IMstuck I tried it and doesent work

Answer 12

yeah that should work

Answer 13

Zrander, as you tried to find the answer, I would love to see what you tried yet. And will be better if I recognize the mistake in your question.

Answer 14

\[\sin2x+\sin x=0\\
2\sin x\cos x+\sin x=0\\
\sin x(2\cos x+1)=0\]
\[\sin x=0~~~~\text{and}~~~~\cos x=-\frac{1}{2}\]

Answer 15

If you have sin(2x)= -sin(x), then what? where do you go with it? Even if you use the fact that sin(2x) = 2sinxcosx, you get c0s(x) = -1/2, not positive 1/2. I think something is wrong.

Answer 16

Or you can just follow this :

\(\sin (2x) = -\sin (x)\)

\(2\sin (x) \cos (x) = - \sin (x) \)

\(2\cos (x) = - 1\) or \(\sin (x) =0 \)
\(\cos (x) = \cfrac{-1}{2} \) or \(\sin (x) = 0 \)

Answer 17

Find the exact solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
sin 2x + sin x = 0

This is the question word for word. Idk. I just dont get it

Answer 18

The cosine of x is -1/2 at 2pi/3 and 4pi/3, not pi/3 and 5pi/3 as zrander23 says they are. It just doesn't come out that way.

Answer 19

the solutions to the problem above are 0, pi, 2pi/3, 4pi/3

Answer 20

This is one of the practice ones on my homework, as i stated earlier, I was just showing how they wanted it formatted
sin 2x − sin x = 0

0, pie/3, pie, 5pie/3

Answer 21

Use the unit circle. Are you allowed to use a unit circle to find the answers once you solve for x?

Answer 22

yeah

Answer 23

Ok, I think I get it, what you're saying. You're saying that the 0, pi/3, pi, 5pi/3 are just EXAMPLES of the format of the answers you find?

Answer 24

yes haha

Answer 25

Ok. Then here we go with solving this mess! ; )

Answer 26

Thoose asnwers you gave me worked, i somewhat understand now!

Answer 27

I looked at my unit circle haha

Answer 28

Do you want to understand thoroughly or do you just want to somewhat understand? I'm a high school math teacher and somewhat understanding does not usually work out too well from my experience! Let us help you if you need to understand better. Do you know how to solve the equations to find what x is?

Answer 29

I give you credit for being a math teacher!, not really

Answer 30

i dont know how to solve it

Answer 31

I struggle with math and this is my last math class i need to talk for college, for my degree

Answer 32

You want help? I would love to help you!

Answer 33

Ill send you a message

Answer 34

Yeah I mean I could use help, for my online class I have one more test and a final

Answer 35

Ok the first thing you have to do is realize that sin(2x) is a double angle. In order to deal with it, you need to use its equivalency and get rid of it. Are you familiar with what the identity of sin(2x) is?

Answer 36

2 (sin x )(cos x ) right?

Answer 37

Yes. So just replace the sin(2x) with 2sinxcosx to get 2sin(x)cos(x) + sin(x) = 0. Now you have a situation where you can simplify by factoring. You see there is a sin(x) on both sides of the " + " sign, so you can factor one out. What do you get when you do that?

Answer 38

2sin(x)cos(x) + sin(x) = 0 becomes \[\sin x(2\cos x +1)=0\]

Answer 39

i was working it on paper and thats what i just got

Answer 40

Ok, that is a perfect start! the sin(x) = 0 is an easy one. So let's deal with the 2 cos(x) + 1 = 0 first. You solve this like any "normal" factor that is set equal to zero. You eventually want to solve for x. So you subtract 1 from both sides and get 2cos(x) = -1. Then to isolate the cos(x) you divide both sides by -2. Now what you have is \[\cos (x) = -\frac{ 1 }{ 2 }\]Are you with me so far?

Answer 41

im following

Answer 42

it makes sense when i write it down on paper, its hard when its an online class based on powerpoints

Answer 43

so basically your breaking it down

Answer 44

A second version of a Mathematica 9 solution is attached.