The answer is {x in RR : x=kpi or x=(2pi)/3+2kpi or x=-(2pi)/3+2kpi, k in ZZ}
Remember:
sin(4x)=2sin(2x)cos(2x)
sin(3x)=sin(x)cos(2x)+cos(x)sin(2x)
sin(2x)(1+2cos(2x))=sin(2x)cos(x)+cos(2x)sin(x)
Remember: 1=2-1
sin(2x)(2+2cos(2x)-1-cos(x))=cos(2x)sin(x)
Remember: 1+cos(2x)=2cos^2(x)
2sin(x)cos(x)(4cos^2(x)-1-cos(x))=cos(2x)sin(x)
We have one set of solutions: A={x in RR :sin(x)=0}={kpi}_(k in ZZ)
We now can simplify
2cos(x)(4cos^2(x)-1-cos(x))=cos(2x)
Remember: cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1
8cos^3(x)-2cos(x)-2cos^2(x)=2cos^2(x)-1
y=cos(x)
8y^3-4y^2-2y+1=0
We notice it's
(2y)^3-(2y)^2-(2y)+1=0
So 2y=-1 and it's the sole solution in RR (we can divide with Ruffini's rule, or calculate the local minimum to prove it)
So we have another set of solutions B={x in RR : cos(x)=-1/2}={(2pi)/3+2kpi} cup {-(2pi)/3+2kpi}
We know we don't have any other solution so the set is A cup B