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Solve: sin(2x)=sinx
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x=2npi n is an integer
2x=x
2x-x=0 1x=0 x=0
double angle formula for sin2x: sin2x=sinx 2sinxcosx=sinx sinx=0 or cosx=1/2 x=0,2pi,pi/3,5pi/3
largrange includes other solutions so he is right
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we have 2sinxcosx=sinx 2sinxcosx-sinx=0 sinx(2cosx-1)=0
sinx=0 or 2cosx-1=0
myininaya is also right, her answer is very thorough
\[\sin(2x)=2\sin(x)\cos(x)\] \[\sin(2x)=\sin(x) \iff 2\sin(x)\cos(x)=\sin(x) \iff 2\cos(x)=1 \iff \cos(x)=\frac{1}{2}\] \[\iff x=\frac{\pi}{3} + 2k \pi, k \in \mathbb{Z} \ or x=-\frac{\pi}{3}+2k \pi, k \in \mathbb{Z}\]
x=2npi, n is an integer cosx=1/2 when x=pi/3+2npi and x=5pi/3+2npi
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