Posted by Marie on Friday, August 3, 2012 at 3:50pm.
sin(2x)=2sin(x)cos(x)
so
sin(2x)+sin(x)=0
=>
2sin(x)cos(x)+sin(x)=0
sin(x)[2cos(x)+1]=0
sin(x)=0 or cos(x)=-1/2
sin(x)=0 => x=0, x=π for 0≤x<2π
cos(x)=-1/2 => x=2π/3 or x=4π/3
Combining, the solution set
S={0,2π/3,π 4π/3}
Add x=2π if the question actually asked for
0≤x≤2π
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