Question

Find all solutions in the interval [0, 2π).

sin2 x + sin x = 0

Answer 1

@SolomonZelman ?

Answer 2

please use this identity:
\[\sin(2x)=2 \sin x \cos x\]

Answer 3

so 2sinxcosx+sinx =0 ?

Answer 4

no exactly, you forgot the `sin x` that you have in your equation already.

Answer 5

ok! now factorise, putting in evidence sin (x), please

Answer 6

im not sure what that means

Answer 7

I write:
\[\sin x(2\cos x+1)=0\]

Answer 8

here is what you did.
\(\large\color{black}{ \color{red}{ \sin(2x)}- \sin x = 0}\)
\(\large\color{black}{ \color{red}{ 2\sin(x)\cos(x)}~~~~~ = 0}\)

but you left something out, didn't you?

Answer 9

okay i see, now what?

Answer 10

first, write the equation that you are up to correctly, please.

Answer 11

you need the +sinx after the cos(x) @SolomonZelman

Answer 12

now, please you have to apply the Law of cancellation of the product

Answer 13

so you got two equations, namely:
sin x=0, and 2 cos x +1=0

Answer 14

ohh it is plus, not minus. so yes and you would be getting,
\(\large\color{black}{ \color{red}{ \sin(2x)}+ \sin x = 0}\)
\(\large\color{black}{ \color{red}{ 2\sin(x)\cos(x)}+\sin(x)= 0}\)

Answer 15

then, factor out of sin(x), since it is the common factor in both terms.

Answer 16

\[\sin x=0, 2 \cos x+1=0\]

Answer 17

so cosx= 1/2?

Answer 18

no, cos x=-1/2, because you have to add -1 to both sides of the second equation first

Answer 19

\(\large\color{black}{ 2\color{blue}{\sin(x)}\cos(x)+\color{blue}{\sin(x)}= 0}\)
\(\large\color{black}{ \color{blue}{\sin(x)}(2\cos(x)+1)= 0}\)

Answer 20

So, either \(\large\color{black}{ \color{blue}{\sin(x)} =0}\) or \(\large\color{black}{ 2\cos(x)+1= 0}\)

Answer 21

okay so then do i look at the unit circle or is there another step?

Answer 22

well, for the second option you need a cos(x) that is equal to -1/2...

Answer 23

please, there are no steps

Answer 24

so is it 0, 2pi/3, 4pi/3?

Answer 25

that's right!, there is another solution, namely x= pi, which derives from sin (x)=0

Answer 26

now i am confused because that is not one of the answer choices? let me list the choices really quick

Answer 27

0, pi, pi/3, 5pi/3
0, pi, pi/3, 2pi/3
0, pi, 4pi/3, 5pi/3
0, pi, 3pi/2

Answer 28

please note that:
x=0, pi are the solution of sin (x)=0, whereas,
x=2pi/3, 4pi/3 are the solutions of 2 cos(x)+1=0

Answer 29

sorry , x=2pi/3, 5pi/3 are the solutions to the equation 2 cos(x)+1=0

Answer 30

so what do i do?

Answer 31

sorry it I wrote before, namely solutiuons are:
x= 0, pi, 2pi/3, 4pi/3

Answer 32

but those arent an answer i can choose

Answer 33

I see, nevertheless those solutions are right

Answer 34

i know but i dont know how to answer it because its an online class so i cant really fight that the answer choices are incorrect...

Answer 35

you can check the correctness by substituting those solutions in your equation

Answer 36

how do i type in the sin^2 part

Answer 37

why sin^2?

Answer 38

okay when i plugged it in it says the only ones that equal 0 are the last choices

Answer 39

sorry is your equation this?
\[(\sin x)^{2}+\sin x=0\]

Answer 40

its sin^2 x

Answer 41

is there a way you can help me with another question?

Answer 42

thn yur equation is that I wrote above,
yes!

Answer 43

ill post it in another question and tag you

Answer 44

you ave to factor oout sin(x), so your equation will ecome:
\[\sin x(\sin x +1)=0\]

Answer 45

so solution s are:
x=0, pi fro equation sin x=0, whereas
x0 3pi/2 from equation sin x+1=0
so the last option is your answer

Answer 46

@adekker