Note first that
#\cos(x+\pi)=-\cos(x)#
#\cos(2(x+\pi))=\cos(2x)#
#\cos(3(x+\pi))=-\cos(3x)#
So the function #f(x)=\cos(x)\cos(2x)\cos(3x)# has a period #\pi#, and its solutions for a value of #1/4# will follow suit.
Using the sum product relation for cosines:
#\cos(x)\cos(3x)=(1/2)(\cos(3x-x)+\cos(3x+x))#
#=(1/2)(\cos(2x)+\cos(4x))#
So then
#\cos(x)\cos(2x)\cos(3x)=1/4=(1/2)(\cos(2x)+\cos(4x))#
#2(\cos(2x)+\cos(4x))=1#
Then from the double angle formula:
#cos(4x)=2\cos^2(2x)-1#
#2(\cos(2x)+2\cos^2(2x)-1)=1#
#4\cos^3(2x)+2\cos^2(2x)-2\cos(2x)-1=0#
Note that the equation comes out with trigonometric functions of #2x# because of the period being #\pi#.
We can factor the zero polynomial:
#(2\cos(2x)+1)(2\cos^2(2x)-1)=0#
So either #\cos(2x)=-(1/2)# or else #\cos(2x)=\pm(\sqrt(2)/2)#. This leads to six sets of solutions having period #\pi#:
If #\cos(2x)=-(1/2)# then #x=\pm(\pi/3)+n\pi#, for any integer #n#.
If #\cos(2x)=((\sqrt(2))/2)# then #x=\pm(\pi/8)+n\pi#, for any integer #n#.
If #\cos(2x)=(-(\sqrt(2))/2)# then #x=\pm((3\pi)/8)+n\pi#, for any integer #n#.
