Number of solution(s) satisfying the equation 1sin⁡x−1sin⁡2x=2sin⁡4x in [0,4π] is equal to

1sin⁡x−1sin⁡2x=2sin⁡4x⇒ sin⁡2x−sin⁡xsin⁡xsin⁡2x=22sin⁡2xcos⁡2x⇒ 2sin⁡2xcos⁡2x−2sin⁡xcos⁡2x=2sin⁡x⇒ sin⁡4x−sin⁡3x+sin⁡x=2sin⁡x⇒ sin⁡4x=sin⁡3x+sin⁡x⇒ 2sin⁡2xcos⁡2x=2sin⁡2xcos⁡x⇒ 2x=2nπ±x   (as x=2kπ is not in domain) ⇒ x=2nπ3,n∈I ( as sin⁡2x≠0)⇒n=1,2,4,5,….Thus, four solutions are 2π3,4π3,8π3,10π3