Core 3 Trig identities, help, anyone, please? Watch

perfectimperfection
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#1
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(C1 has the equation) y = cos2x - 2sin^2 x
(C2 has the equation) y = sin2x

show that the x coordinates of the points of intersection of C1 and C2 satisfy the equation 2cos2x - sin2x = 1

so i set C1 and C2 equal to each other and then basically cos i cant see a clear cut way of doing it, i tried using most identities and angle formulae and such to mkae iy work, but with no results. Have tried loads of method none seem to work.

It's only worth 3 marks so cant possibly take that much effort but im stuck!

Thanks to anyone who helps in advance!
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DKer
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Try using cos2x=1-2sin^2x
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perfectimperfection
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(Original post by DKer)
Try using cos2x=1-2sin^2x
then you get 1 - 2sin^2x - 2sin^2x = sin2x
1 - 4sin^2 x = sin2x
i have tried doing that but up to here get confused, or am i doing something blatently wrong?
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DKer
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Your doing it good as far as I know.

Your equation
1 - 4sin^2 x = sin2x
is solvable, its really just a quadratic now, maybe it might look nicer in this form:

4sin^2u+sinu-1=0 where u=2x, so you just need to half your solutions.

See how that goes, I'll warn you tho there may well be better ways to do it, than what I'm suggesting.
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