All you need is the double angle identity:
[tex]\cos^2x=\dfrac{1+\cos2x}2[/tex]
So we have
[tex]\cos^4x=(\cos^2x)^2=\left(\dfrac{1+\cos2x}2\right)^2=\dfrac{1+2\cos2x+\cos^22x}4[/tex]
Apply the identity again to the squared term:
[tex]\cos^4x=\dfrac{1+2\cos2x+\frac{1+\cos4x}2}4=\dfrac{3+4\cos2x+\cos4x}8[/tex]
