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Mathematics 12 Online
OpenStudy (anonymous):

thanks

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

\[2\sin^2(x)=\cos(x)\]?

OpenStudy (misty1212):

start by writing all in terms of cosine i.e use \[\sin^2(x)=1-\cos^2(x)\] then you will have a quadratic equation to solve in terms of cosine only

OpenStudy (anonymous):

hey

OpenStudy (misty1212):

you got that ?

OpenStudy (misty1212):

start with \[2(1-\cos^2(x))=\cos(x)\]

Directrix (directrix):

Is this: 2sin^2x=2 c+osx 2sin^2x=2 cosx or something else

Directrix (directrix):

or, 2sin^2x=2 + cosx

OpenStudy (anonymous):

the second one @dir

Directrix (directrix):

2 sin² (x) = 2 + cos (x) This is not the same one that Misty used. Maybe you could check your text one more time to see if this is it. Also, I don't see any instructions. @cutegirl

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