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Solve : cos3x + sin2x – sin4x = 0

Solution :

we have, cos3x + (sin2x – sin4x) = 0

= cos3x – 2sinx.cos3x = 0

  (cos3x)(1 – 2sinx) = 0

  cos3x = 0  or  sinx = 12

  cos3x = 0 = cosπ2  or  sinx = 12 = sinπ6

  3x = 2nπ ± π2  or  x = mπ + (1)mπ6

  x = 2nπ3 ± π6  or  x = mπ + (1)mπ6; (n, m I)


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