Question

Help me solve: sin^2x-sinx=0

Answer 1

0 and 1
sin x=0 and 1

Answer 2

npi, pi/2+n2pi

Answer 3

\[\sin^2x-\sin x=0\]
\[\sin x( \sin x-1)=0\]
\[\frac{\sin x( \sin x-1)}{\sin x}=\frac{0}{\sin x}\]
\[ \sin x-1=0\]
\[ \sin x=1\]
\[\sin^{-1} (1)=x\]
This occurs at x=90

Answer 4

\[\sin x=0\]
\[\sin ^{-1}(0)=x\]
This occurs at x=0 and x=360

Answer 5

dang it ill give ya a medal anyways
ARGHHHHHH I HATE YOUUUUUUUUUUUUUUU

Answer 6

sin^2x-sinx=0
let y= sin x
y^2-y=0
y(y-1)=0, y=0 or 1
y=sinx
0=sinx, x=n*pi
1=sinx, x=pi/2+2pi*n

Answer 7

dont forget osscilations

Answer 8

90+n*360. nEall integers

Answer 9

dan what does n stand for?

Answer 10

ohhh for alll cycles

Answer 11

gotcha gotcha

Answer 12

and for every 180 degree angle so
180 + n 180

Answer 13

or npi

Answer 14

lose the habit of using degrees

Answer 15

hahaha ok

Answer 16

fuk u

Answer 17

please dont swear on OS

Answer 18

It is a user friendly site for all ages

Answer 19

hahahahaha

Answer 20

welcome to OS sloan

Answer 21

Im gonna beat u up soon

Answer 22

@sloan16 fell asleep -.-

Answer 23

thank you guys so much

Answer 24

You are welcome sloan

Answer 25

give a medal to swissgirl

Answer 26

dan shut up -.-