Soru tam olarak $\sqrt{4cos4x+16cos2x+12}=?$ şeklinde mi?
Evet..................
$\sqrt{4(2cos^22x-1)+16cos2x+12}=\sqrt{8cos^22x+16cos2x+8}$
$=\sqrt{8(cos2x+1)^2}=2\sqrt2|cos2x+1|=2\sqrt2cos2x+2\sqrt2$ dir.