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

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

Mathematics
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sin2x=2sinxcosx so 2sinxcosx=sinx 2sinxcosx-sinx=0 sinx(2cosx-1)=0 sinx=0 and 2cosx-1=0
lets solve sinx=0 first
what x makes sinx zero

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Other answers:

think about your unit circle
sin(0) is what?
0 or pi
sin(0)=0 sin(2pi)=0 so x=0 and 2pi now we have the other equation 2cosx-1=0 => cosx=1/2
what can plug in where the x is that give you 1/2 (we are looking at the equation cosx=1/2)
60 degrees
the other btw was 0 degrees or 180 degrees for sin
oh yes and pi very good for the first
so answeres are 0,1/2,pi?
so 60 degrees and what else for this equation equation
no not 1/2
60 degrees or pi/3 there is also another one for the second equation
ah
0,pi/3,pi that seems like all we can get i think cause thats both sin and hte one cos
@myininaya: Congrats for the new title. I was you 100th fan by the way :D
i was 99th XD
thanks lol
so anyways what is cos(300 degrees)
10pi/6
no maybe it will help if i wrote it in radians what is cos(5pi/3)
isnt that the same just reduced?
ok 10pi/6=5pi/3 but what is cos(5pi/3)
=1/2
-1/2?
isnt it - cause bottom right quadrent?
nvm cos is x so positive if on right side
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
you have 5 solutions in all
thank you
np

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