#sin(2x)=1#.
In general, equations of the form #sin(theta)=K# and solved by #theta = arcsin(K)+2npi# and #pi-theta=arcsin(K)+2npi#, #n in ZZ#, because of the periodicity of the sine function#.
Note that #arcsin(1)=pi/2#. This means that if #sin(2x)=1#,
#2x=pi/2+2npi#,
#x=pi/4+npi#.
#pi-2x=pi/2+2npi#,
#2x=pi/2-2npi#,
#x=pi/4-npi#.
As #n in ZZ# these solution sets are identical.
So if #sin(2x)=1# then #x=pi/4+npi#, #n in ZZ#.
Then
#cos(8(pi/4+npi)+2cos(6(pi/4+npi))+cos(4(pi/4+npi))#,
#cos(2pi+4(2npi))+2cos((3pi)/2+3(2npi))+cos(pi+2(2npi))#.
Multiplies of the periodicity can be removed and the values substituted to give,
#1+2*0-1=0#.
In this case, it was safe to assume #sin(2x)=1 => x=pi/4#. This will not always be the case! So it is good to solve more generally.
