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|dw:1359396748498:dw|

what do i do one i've done this

when I do this :
cos^2x-sin^2x=0
cos(2x)=0
cosx=0
x=pi, 2pi, +2k ?

JamesJ i feel like i did this wrong

|dw:1359397774601:dw|

If
cos(2x) = 0
then
2x = pi/2 + k.pi, for any integer k.
Hence
x = pi/4 + k.pi/2, for any integer k.

You can see those solutions on this graph as well.

Thank you all it's starting to make a lot more since to me now so I just gotta know my identities?

Yes. That is key.