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#
