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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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