Can someone please check my work?

Find the sum of the roots of sin2x = sin x in the interval [0, 2pi].

Okay, so sin2x = 2sinx cosx right?
Then I have 2sinx cosx = sin x
cos x = sin x / 2 sin x
cos x = 1/2
x = pi/3, 5pi/3
Therefore the sum is 2pi. My book says it's 5pi, I don't get it!

Question

Can someone please check my work?

Find the sum of the roots of sin2x = sin x in the interval [0, 2pi].

Okay, so sin2x = 2sinx cosx right? 
Then I have 2sinx cosx = sin x
cos x = sin x / 2 sin x
cos x = 1/2
x = pi/3, 5pi/3
Therefore the sum is 2pi. My book says it's 5pi, I don't get it!

anonymous · Answer

don't cancel the sinx like that... what you'll need to do is factor...

dumbcow · Answer

dont divide by sin x 

--> 2sinx cosx - sinx = 0
--> sinx (2cosx -1) = 0
sinx = 0 --> x = 0,pi
cos x = 1/2 --. x = pi/3, 5pi/3

anonymous · Answer

2sinx cosx = sin x
2sinxcosx - sinx = 0

factor sinx...

dumbcow · Answer

oh i didn't see interval...2pi is also a solution

anonymous · Answer

notice if you cancelled the sinx, you'd be missing x=0,pi....

anonymous · Answer

oh, sinx=0
x=pi
x=2pi

anonymous · Answer

Thank you both! Much appreciated