Hi I have tried to solve this... (but my teacher never gets satisfied).

Solve the trigonometric equation:

4sinx*sin2x*sin4x=sin3x

My solution so far...

4sinx*sin2x*sin4x−sin3x=0

3x=sinx(3-4((sinx)^2)

0=4sinx*sin2x*sin4x-sinx(3-4*((sinx)^2)= sinx(4sin(2x)) *sin4x-(3-4*((sinx)^2)

So:

0=sinx

or

0=4sin2x*sin4x-(3-4*((sinx)^2)

Please help.