Precalculus, Chapter 5, 5.5, Section 5.5, Problem 7

sin(2x)-sin(x)=0 , 0<=x<=2pi
using the identity sin(2x)=2sin(x)cos(x)
sin(2x)-sin(x)=0
=2sin(x)cos(x)-sin(x)=0
=(sin(x))(2cos(x)-1)=0
solving each part separately,
sin(x)=0 or 2cos(x)-1=0
General solutions for sin(x)=0 are,
x=0+2(pi)n, x=pi+2(pi)n
solutions for the range 0<=x<=2pi are,
x=0 , x=pi , x=2pi
Now solving 2cos(x)-1=0,
2cos(x)=1
cos(x)=1/2
General solutions for cos(x)=1/2 are,
x=(pi)/3+2pin , x=(5pi)/3+2pin
solutions for the range 0<=x<=2pi are,
x=pi/3 , x=(5pi)/3
Hence all the solutions are,
x=0,pi,2pi,pi/3 , (5pi)/3