# math

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Solve sin2x + sin4x = cos2x + cos 4x for x in the interval[0,2pi)
Hint: the following substitution should come in handy: sin3x= cos3x . tan3x

• math -

sin2x + sin4x = 2sin3x cosx
cos2x + cos4x = 2cos3x cosx
so, we have

sin3x cosx = cos3x cosx
tan3x = 1
so, 3x = pi/4 or 5pi/4
x = pi(1+4k)/12 for all 0<=k<=5
also x=k*pi/2 (where cosx = 0)

• math -

cos4x-cos2x=sin4x-sin2x

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