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Solve cos2x-cos4x=sin2x
By Anonymous on Thursday, June 17, 1999 - 10:33 am:

I can't rearrange this to find its general solution

cos2x-cos4x=sin2x

So far I've used Rcos(x+a) to get

sqrt2cos(2x+45)=cos4x

but I don't know if this is correct and if so where to go from there.

I would be very grateful if anyone could help.

By Alex Barnard (Agb21) on Monday, June 21, 1999 - 05:00 pm:

Actually the best thing to do is to use trig identities to combine the two cos terms.

cos(3x-x)=cos(3x)cos(x) + sin(3x)sin(x)
cos(3x+x)=cos(3x)cos(x) - sin(3x)sin(x)

=> cos(2x)-cos(4x) = 2sin(3x)sin(x)

Now use sin(2x) = 2sin(x)cos(x) to get:

sin(3x)sin(x) = sin(x)cos(x)

So either sin(x)=0 [ie. x=0 or 180 degrees]

Or sin(3x)=cos(x).

Now cos(x) = sin(x+90)

So solve sin(3x) = sin(x+90) [ie. x=45 or 225 degrees]

Hope that helps,

AlexB.