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.