giải pt: (cos2x+3cotx+sin4x)/ (cot2x-cos2x ) =2

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1. (Cos2x+ 3cot2x +sin4x)/ (cot2x -cos2x ) = 2

   ⇒ ( cos2x.sin2x + 3cos2x + sin4x.sin2x)/sin2x = 2(cos2x – cos2x.sin2x)/sin2x

   ⇒ (1/2sin4x + 3cos2x + 1/2(cos2x – cos6x) = 2 (cos2x – 1/2sin4x)

   ⇒ 3/2sin4x + 3/2cos2x – 1/2cos6x =0

   ⇒ 3sin4x + 3cos2x – cos6x =0

   ⇒ 3sin4x + 2cos2x + cos2x – cos6x =0

   ⇒ 3sin4x + 2cos2x -2sin4x . sin2x =0

   ⇒ cos2x( sin2x + 1 – 2 sin^2(2x))=0

   ⇒ cos2x=0 hay sin2x + 1 – 2sin^2(2x) =0 => sin2x = -1/2 hay sin2x =1

   … Còn lại bạn tự làm nha …

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