Ta có: $y=\dfrac{(1-\cos2x)^4}{8} +\cos^42x$
Đặt: $\cos2x=t, t \in [-1;1]$
Xét hàm: $f(t)=\dfrac{(1-t)^4}{8} +t^4$
Ta có: $f'(t)= -\dfrac{1}{2}(1-t)^3 +4t^3$
$f'(t)=0 \Leftrightarrow 2t=1-t \Leftrightarrow t=\frac{1}{3}$
Lập bảng biến thiên ta có:
$\max_{t\in[-1;1]}f(t)=f(-1)=3$
$\min_{t\in[-1;1]}f(t)=f(\dfrac{1}{3})=\dfrac{1}{27}$