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Mathematics 19 Online
OpenStudy (brydenwright):

2 ((sin^2)x) = sin x

OpenStudy (loser66):

subtract sin(x) both sides, what do you get?

OpenStudy (brydenwright):

2 ((sin^2)x) - sin x = 0

OpenStudy (brydenwright):

But how would I factor?

OpenStudy (evoker):

pull out the common factor

OpenStudy (sshayer):

\[2\sin^2x-\sin x=0,\sin x \left( 2\sin x-1 \right)=0\] either sinx=0 x=? or \[\sin x=\frac{ 1 }{ 2 },find x\]

OpenStudy (mathmale):

Regarding your "2 ((sin^2)x) = sin x" ... I think you probably mean 2 (sin x)^2 = sin x Think about this. What's your conclusion? If you agree with my interpretation, you'll then have \[2 (\sin x)^2=\sin x, \]

OpenStudy (mathmale):

or \[2 \sin^2x=\sin x\]

OpenStudy (mathmale):

How would you solve that for x?

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