2x = 0,π,2π,3π       
 x = 0,π/2,π,3π/2    
                     2x = π/6,11π/6,7π/6,23π/6  
                      x = π/12,11π/12,7π/12,23π/12
8 solutions.

Edwin

Your starting equation is


    sin(4x) − sqrt(3)*sin(2x) = 0


Substitute (replace) here     sin(4x) = 2sin(2x(*cos(2x).


    2sin(2x)*cos(2x) - sqrt(3)*sin(2x) = 0.


Factor left side


    2sin(2x)*(cos(2x)-sqrt(3)) = 0.


This equation deploys in two equations


    a)  sin(2x) = 0.


        It has the roots  2x = 0, ,  ,    in the interval  [0,)

        that create the roots

        x = 0,  ,  ,    in the interval  [0,).



    b)  cos(2x) = .

        It has the roots  2x = ,  ,  ,    in the interval  [0,)

        that create the roots

        x = ,  ,  ,    in the interval  [0,).


ANSWER.  There are 8 roots  x = 0,  ,  ,  

                                ,  ,  ,    in the interval  [0,).