The equation is
#cos^3x+cos^2x-4cos^2(x/2)=0#
Let #u=x/2#
#cos^3 2u+cos^2 2u-4cos^2(u)=0#
But
#cos2u=2cos^2u-1#
Therefore,
#(2cos^2u-1)^3+(2cos^2u-1)^2-4cos^2u=0#
#8cos^6u-12cos^4u+6cos^2u-1+4cos^4u-4cos^2u+1-4cos^2u=0#
#8cos^6u-8cos^4u-2cos^2u=0#
#2cos^2u(4cos^4u-4cos^2u-1)=0#
Therefore,
#{(cos^2u=0),(4cos^4u-4cos^2u-1=0):}#
#<=>#, #{(cosu=0),(cos^2u=(4+sqrt(16+16))/8),(cos^2u=(4-sqrt(16+16))/8):}#
#<=>#,
#u=pi/2+2kpi#
# u=3/2pi+2kpi#
#cosu=+-sqrt((1+sqrt2)/2)#
#cosu=+-sqrt((1-sqrt2)/2 ) #
#u=arccos(+sqrt((1+sqrt2)/2))#
#=>#, #u=O/# as #cosu>1#
#u=arccos(-sqrt((1+sqrt2)/2))#
#=>#, #u=O/# as #cosu<-1#
#u=arccos(+sqrt((1-sqrt2)/2))#
#u=arccos(-sqrt((1-sqrt2)/2))#
Finally,
#x/2=pi/2+2kpi#, #=>#, #x=pi+4kpi#
#x/2=3pi/2+2kpi#, #=>#, #x=3pi+4kpi#
#x/2=O/#, #=>#, #x=O/#
#x/2=arccos(+sqrt((1-sqrt2)/2))+2kpi#, #=>#, #x=2arccos(+sqrt((1-sqrt2)/2))+4kpi#
#x/2=arccos(-sqrt((1-sqrt2)/2))+2kpi#, #=>#, #x=2arccos(-sqrt((1-sqrt2)/2))+4kpi#
