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Solve equation on the interval [0,2pi) of sinx=sin(2x)
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first of all, of course it happen at x=0, so now assume \[x\neq 0\] : \[\sin(x)=\sin(2x)=\sin(x)\cos(x)\] so we have \[\cos(x)=1/2\] , where does that happen in your interval?
I meant \[\sin(2x)=2\sin(x)\cos(x)\]
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