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September 1, 2014

September 1, 2014

Posted by **Marie** on Friday, August 3, 2012 at 3:50pm.

- Trigonometry -
**MathMate**, Friday, August 3, 2012 at 6:06pmsin(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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