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Answer 1 · 2017-11-24T14:59:22

#4cosx cos2x cos3x - 2cos4x =1#

#=>4cosx cos3x cos2x - 2cos4x =1#

#=>(2cos4x+ cos2x )cos2x - 2cos4x =1#

#=>2cos4xcos2x+ 2cos^2 2x - 2cos4x =1#

#=>2cos4xcos2x+ 1+cos4x - 2cos4x =1#

#=>2cos4xcos2x-cos4x =0#

#=>cos4x(2cos2x-1) =0#

when

#cos4x=0#

#=>4x=(npi)/2" where " n in ZZ#

#=>x=(npi)/8" where " n in ZZ#

When

#2cos2x-1=0#

#=>cos2x=1/2=cos(pi/3)#

#=>2x=2mpipmpi/3" where " m in ZZ#

#=>x=mpipmpi/6" where " m in ZZ#