Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Find all solutions in the interval [0, 2π). 2 sin2x = sin x

OpenStudy (anonymous):

sin^2x*

OpenStudy (mertsj):

Subtract sin x from both sides and factor out sin x. Then set each factor equal to 0 and solve.

OpenStudy (anonymous):

Okay, thanks. So you factor and set it to 0?

OpenStudy (mertsj):

yes. Set each factor to 0

OpenStudy (anonymous):

I'm sorry, I had a blank moment, how do you factor? Once again, so sorry about the stupid question.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Okay, I did it correctly, thought I did it wrong. :P

OpenStudy (anonymous):

Thanks to the both of you.

OpenStudy (mertsj):

yw

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!