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Mathematics 22 Online
OpenStudy (anonymous):

Find all solutions to the equation in the interval [0, 2pi): 2sin^2x=sinx I have no idea where to start! Please help!

OpenStudy (anonymous):

\[2\sin^2x=sinx\] \[2\sin^2x-sinx=0\] \[\sin x(2sin x-1)=0\]

OpenStudy (anonymous):

Yes! Thank you so much!

OpenStudy (anonymous):

now can u find the value of x?

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