Question

if trying to find all solutions on the interval 0,2pi
sin2x=sinx
so x=1/2? or how do i solve?

Answer 1

sin2x=2sinxcosx
so 2sinxcosx=sinx
2sinxcosx-sinx=0
sinx(2cosx-1)=0
sinx=0 and 2cosx-1=0

Answer 2

lets solve sinx=0 first

Answer 3

what x makes sinx zero

Answer 4

think about your unit circle

Answer 5

sin(0) is what?

Answer 6

0 or pi

Answer 7

sin(0)=0
sin(2pi)=0
so x=0 and 2pi
now we have the other equation 2cosx-1=0 => cosx=1/2

Answer 8

what can plug in where the x is that give you 1/2 (we are looking at the equation cosx=1/2)

Answer 9

60 degrees

Answer 10

the other btw was 0 degrees or 180 degrees for sin

Answer 11

oh yes and pi very good for the first

Answer 12

so answeres are 0,1/2,pi?

Answer 13

so 60 degrees and what else for this equation equation

Answer 14

no not 1/2

Answer 15

60 degrees or pi/3
there is also another one for the second equation

Answer 16

ah

Answer 17

0,pi/3,pi that seems like all we can get i think cause thats both sin and hte one cos

Answer 18

@myininaya: Congrats for the new title. I was you 100th fan by the way :D

Answer 19

i was 99th XD

Answer 20

thanks lol

Answer 21

so anyways what is cos(300 degrees)

Answer 22

10pi/6

Answer 23

no maybe it will help if i wrote it in radians
what is cos(5pi/3)

Answer 24

isnt that the same just reduced?

Answer 25

ok 10pi/6=5pi/3
but what is cos(5pi/3)

Answer 26

=1/2

Answer 27

-1/2?

Answer 28

isnt it - cause bottom right quadrent?

Answer 29

nvm cos is x so positive if on right side

Answer 30

k :)
so the solutions to sinx=0 in the interval [0,2pi] are
0,pi,2pi
and the solutions to cosx=1/2 in the interval [0,2pi] are
pi/3, 5pi/3

Answer 31

you have 5 solutions in all

Answer 32

thank you

Answer 33

np