Question

Find all solutions to the equation.
sin2x+sinx=0

Answer 1

is it sin^2x ?

Answer 2

\[\sin^2x+sinx=0\]

Answer 3

alright that's the quadratic equation factor t

Answer 4

what is the common factor ?

Answer 5

sin?

Answer 6

right take it out

Answer 7

it's sin(x) so when we take out sinx from sinx what should we get ?

Answer 8

x represent theta here

Answer 9

\[sinx^2(2)\]

Answer 10

hmm \[\rm \sin^2x +sinx=0\] can be written as \[\large\rm \sin(x) *\sin(x)+\sin(x)=0\]
now take out sinx from both terms

Answer 11

sin(x) is common factor

Answer 12

There is no way to take it out if they are all sin(x)

Answer 13

Or does this mean the solution is all real numbers?

Answer 14

hmm lets make it simple
sin(x) = y
\[\huge\rm y^2+y=0\] take out the common factor

Answer 15

let sin(x) = y^^^

Answer 16

...lol... am I trying to get y=

Answer 17

Im sorry

Answer 18

we have to find factor of that quadratic equation

Answer 19

what's the common factor ?
y^2+y=0 ?

Answer 20

Y

Answer 21

y*y+y=0

Answer 22

right now take out y from both terms
from y^2 and from y what would you get
in other words when we take out common factor we basically divide both terms by common factor

Answer 23

y(y+1)=0

Answer 24

correct now set it equal to zero
\[\rm y=0 ~~~~~~~~~~~~y+1 =0\]
now we can replace it back with sin(x)
\[\sin(x)=0~~~~~~~~~~~~~~~~\sin(x)+1=0\]

Answer 25

Okay I see what you mean

Answer 26

make sense ?
sin(x) is one term not just sin
\[\sin x \rightarrow \sin \theta \]

Answer 27

Understood

Answer 28

alright sin(x)=0 and sin(x)+1 = 0
^^^ solve for sin(x)

Answer 29

sin(x)=0,-1

Answer 30

correct and then unit circle!

Answer 31

180 and 270?

Answer 32

find `all` solutions

Answer 33

um

Answer 34

\[180 \le x \le 270\]

Answer 35

180 and 270 are correct but there is another angle where sin(x) =0

Answer 36

But that goes on forever

Answer 37

ye true :P

Answer 38

\[180 \le x \le 270\] hmm this make sense

Answer 39

would you rather me say pi and 2pi

Answer 40

well that depends on the statement does it say in radins or degrees?
some times the just give in interval ( pi </ x </2pi)

Answer 41

an*

Answer 42

Find all solutions to the equation.
\[\sin^2x+sinx=0\]

Thats exactly what it gives me

Answer 43

\[180 \le x \le 270\] solutions between180 to 270
but look at the 0/260 degree

Answer 44

sin(x)=0 at 0 degree and we need `all` solutions i'm not sure what they want like radian or degree
i'll go with degree..

Answer 45

gtg and hmm im not sure

Answer 46

@freckles ,-,

Answer 47

what part are you stuck on @kjones0331 ?

Answer 48

What solution to put as the answer because this is a free response question

Answer 49

It says all solutions, be we came up with infinite

Answer 50

solve
\[\sin^2(x)+\sin(x)=0 \text{ where } x \in [180,270] ?\]
this is the question?

Answer 51

First you need to factor sin^2(x)+sin(x)

Answer 52

oh you need all the solutions?

Answer 53

it just says solve sin^2(x)+sin(x)=0?

Answer 54

wait..

Answer 55

ok well you still need to factor sin^2(x)+sin(x)

Answer 56

Could \[x \epsilon, [180,270]\] be the answers

Answer 57

ok so have you factors sin^2(x)+sin(x) yet?

Answer 58

Yes

Answer 59

Up top

Answer 60

ok set both factors equal to 0

Answer 61

and just use unit circle

Answer 62

We did, we got \[sinx=0\] and \[sinx=-1\]

Answer 63

Which on the unit circle could be 180 or 270

Answer 64

ok sin^2(x)+sin(x)=0
sin(x)(sin(x)+1)=0
sin(x)=0 or sin(x)=-1
this should be easy look at the y-coordinates on the unit circle where do you see y is 0 and y is -1 where the angles are between 180 and 270 (inclusive)

Answer 65

and yes exactly 180 or 270 are the solutions

Answer 66

sin(x)=0 when x is 180 deg
sin(x)=-1 when x is 270 deg

Answer 67

How would I write that as an answer though?

Answer 68

sin^2(x)+sin(x)=0 where 180<=x<=270
has solutions x=180 or 270

Answer 69

that is it
we already solved it

Answer 70

So we should leave it in deg? Basically writing:\[180 \le x \le 270\]

Answer 71

you gave the interval in degree the answer should be in degrees

if the interval was in radians then you would answer in radians

Answer 72

But there was no deg or radians in the original equation, that was what Nnesha and I were stuck on. We have the answer, just dont know how to write it.

Answer 73

I thought you said the question was solve sin^2(x)+sin(x)=0 where 180<=x<=270?

Answer 74

is that not the question?

Answer 75

Im sorry for the misunderstanding.. the original problem is what I typed as the question

Answer 76

so the question is just sin^2(x)+sin(x)=0 and there was no restrictions on x?

Answer 77

No

Answer 78

what?

Answer 79

Im sorry, no there are no restrictions

Answer 80

that is what I said and you said no though

Answer 81

\[\sin^2(x)+\sin(x)=0 \\ \sin(x)[\sin(x)+1]=0 \\ \sin(x)=0 \text{ or } \sin(x)=-1 \]
ok I would just put the answer in radians

first find all the solutions in the first rotation [0,2pi]

Answer 82

and then we will just add 2pi*k where k is an integer
to get all the solutions

Answer 83

you actually already found when sin(x)=-1 on [0,2pi]
which is when x=3pi/2
so all the solutions to sin(x)=-1 would be x=3pi/2+2pi*k

now sin(x)=0 on [0,2pi] has solutions x=0,pi,2pi
if you notice the pattern there then you know that all the solutions to sin(x)=0
would be x=k*pi

Answer 84

where k is an integer

Answer 85

So I would answer the question with x=kpi?

Answer 86

that is one solution
you have to also state the other

Answer 87

I am getting 3pi/2 ... 7pi/2 ... 11pi/2 ... and so on

Answer 88

why not just say for the sin(x)=-1 we have x=3pi/2+2pi*k ?

Answer 89

Then 3pi/2 wouldnt fit

Answer 90

why not?

Answer 91

But I could type that seperately

Answer 92

k=0 is an integer

Answer 93

the 2pi

Answer 94

I don't know what to tell you
sin(x)=0 has solutions x=k*pi
sin(x)=-1 has solutions x=3pi/2+2pi*k
where k is an integer

and 3pi/2 does fit in that last one because k=0 is an integer

Answer 95

3pi/2+2pi*(0)
3pi/2+0
3pi/2 <---see 3pi/2+2pi*k does give us 3pi/2

Answer 96

Oh okay, sorry

Answer 97

the set of integers are {...,-4,-3,-2,-1,0,1,2,3,4,....}

Answer 98

Thank you so much