#cos^2(theta) ne (cos^2(theta)+1)/2# in general. There are solutions to the equation, though:

#cos^2(theta) = (cos^2(theta)+1)/2#

#2cos^2(theta) = cos^2(theta)+1#

#cos^2(theta) = 1#

#cos(theta)=1# or #cos(theta)=-1#

#cos(theta)=1 rarr theta = 2pi*n#, #n in ZZ#

#cos(theta)=-1 rarr theta = pi + 2pi*n#, #n in ZZ#.

However! I suspect there's a typo because the following is true:

#cos^2(theta) = (cos(2theta)+1)/2#

This is a power reducing formula that's quite useful and comes from the double angle formula for cosine:

#cos(2theta) = 2cos^2(theta)-1#

If we solve this for #cos^2(theta)#,

#cos(2theta) +1= 2cos^2(theta)#

#(cos(2theta) +1)/2= cos^2(theta)#

Where the double angle formula comes from is a bit of a rabbit hole, so I'm going to stop there.