Trigonometry
posted by Marie .
Solve sin 2x + sin x = 0 for 0 (Less then or equal to) x (Less then or equal to) 2pi.

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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