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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!
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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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