Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

3sinx=1+cos2x

OpenStudy (anonymous):

if i get 16 kills and 6 deaths what is my kd :O

OpenStudy (anonymous):

16:6, so 8:3?

OpenStudy (anonymous):

\[\cos 2x= 1- 2\sin^2x \] so \[3sinx= 1 + 1 - 2\sin^2 x\]

OpenStudy (anonymous):

okay, from where would I go with that?

OpenStudy (anonymous):

\[2\sin^2x+3sinx-2=0\] Then let sinx be u so that you have \[2u^2+3u-2=0\] can you go on from there?

OpenStudy (anonymous):

would I factor it so it ends up becoming (2u-1)(u+2)=0 and then set u back equal to sin^2x? or just sinx?

OpenStudy (anonymous):

well u=sin x so you would set it back to sin x. But yes, the factoring part is right.

OpenStudy (anonymous):

okay so I would get sinx=1/2 and -2? thus, pi/6?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

alright, thank you!

OpenStudy (anonymous):

np

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!