Answer
$\frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{8}cos4x$
Work Step by Step
Use $cos^2x=\frac{1+cos2x}{2}$, we have
$cos^4x=(\frac{1+cos2x}{2})^2=\frac{1}{4}(1+2cos2x+cos^22x)=\frac{1}{4}(1+2cos2x+\frac{1+cos4x}{2})
=\frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{8}cos4x$