sin5(x)=sin(x),
sin5(x)-sin(x)=0,
sin(x)(sin4(x)-1)=0,
sin(x)(sin2(x)-1)(sin2(x)+1)=0,
sin(x)(sin(x)-1)(sin(x)+1)(sin2(x)+1)=0.
So sin(x)=0 is a solution⇒x=nπ where n is an integer.
sin(x)=1 is a solution⇒x=2πn+π/2=½π(4n+1).
sin(x)=-1 is a solution⇒x=2πn+3π/2=½π(4n+3).
The last two solutions can be combined: x=π(2n+1±½).
SOLUTION x=nπ or π(2n+1±½).