Question

Find all solution to the equation
sin^2x+sinx=0

Answer 1

do you know how to factor sin^2(x)+sin(x)?

Answer 2

no i dont

Answer 3

do you know sin^2(x) is the same as sin(x)*sin(x) ?

Answer 4

yes

Answer 5

do you also know sin(x)=sin(x)* 1 ?

Answer 6

yes

Answer 7

\[\sin^2(x)+\sin(x) \\ \sin(x) \cdot \sin(x)+ \sin(x) \cdot 1 \\ \color{red}{\sin(x)} \cdot \sin(x)+\color{red}{\sin(x)} \cdot 1 \]
so do you know how to factor this now?

Answer 8

do i divide by sinx?

Answer 9

a b+ a c=a(b+c)

Answer 10

do you mean divide both sides by sin(x) ?

Answer 11

the only bad thing about that is you may be losing out on some zeros

Answer 12

yea thats what i was thinking but i guess its not correct

Answer 13

or solutions

Answer 14

so i do the a b+ ac=a(b+c) thing?

Answer 15

yep that is how you will factor what I have asked you to factor
that is known as the distributive law/property (don't know if was law or property-can't remember :p )

Answer 16

ok so how do i tell whats a b or c?

Answer 17

a is the the factor a*b and a*c have in common

Answer 18

what do you think a is here?
what factor does sin(x)*sin(x) and sin(x)*1 have in common?

Answer 19

sinx

Answer 20

right

Answer 21

ok so a = sinx so b is what?

Answer 22

is it also sinx?

Answer 23

yep

Answer 24

and c=1?

Answer 25

yep

Answer 26

it is just multiplication
a(b+c)
=
a*b+a*c

Answer 27

ok let me see if i got this

Answer 28

sinx(sinx+1)=Sinx x Sinx + Sinx?

Answer 29

that extra x between the sin(x)'s shouldn't be there
but I don't think you meant to write it there

yes \[
\sin^2(x)+\sin(x)=\sin(x)(\sin(x)+1)\]

Answer 30

so is that the answer?

Answer 31

so \[\sin(x)(sin(x)+1)=0\]
this means we have to solve two equations
\[\sin(x)=0 \text{ or } \sin(x)+1=0\]

Answer 32

ok then we use the unit circle and find where y is 0 on the unit circle?

Answer 33

sounds great
that will work for the first equation sin(x)=0

Answer 34

you will also need to find when y is -1 on the unit circle to solve sin(x)+1=0

Answer 35

ok so the solutions are 0, 180, and 270?

Answer 36

it says find all solutions
the solutions you have given here are solutions in [0,360)

Answer 37

so what are the other solutions?

Answer 38

use the period of the sine function is 360 degrees
this means keep adding 360 to your solutions above
and you will still have a solution to your equation

Answer 39

you can do by adding 360*k to each of your solutions in [0,360)

Answer 40

where k is a integer

Answer 41

so the solutions are 0*k, 180*k, and 360*k. Where k is an integer

Answer 42

the solutions you just told me were 0,180, and 270
I have asked you just to add on +360k to all of these to get all solutions

Answer 43

for example
the first solution you got was 0
just do 0+360k or 360k

Answer 44

ok so solutions are 0+360k, 180+360k, and 270+360k?

Answer 45

yep and you can write the first one as 360k since 0+anything=that anything

Answer 46

ok thank you for your help

Answer 47

np

Answer 48

Please close this question ^^