We want to solve for x
sinx=0 and (4sin2x-1)=0 split into two equations
(2sinx-1)(2sinx+1)=0 factor the second (difference of two squares)
so now we have 3 factors whoas product is 0 so
sinx=1/2, sinx=-1/2 and sinx=0
taking the inverse of sinx=0 gives x=0,π as the solution for sinx=0
next we find the solution of sinx=1/2 or x=π/6, 5π/6,
There are two more solutions to sinx =-1/2 can you find them? When you do you will have found all the solutions to (4sin2x-1)*sinx = 0 in the interval [0,2π)
Jim