Dễ dàng có $\sin^6 x +\cos^6 x = (\sin^2 x +\cos^2 x)^3 -3\sin^2 x \cos^2 x (\sin^2 x +\cos^2 x)=1-3\sin^2 x \cos^2 x$
$=1-\dfrac{3}{4}\sin^2 2x =1-\dfrac{3}{8}(1-\cos 4x)=\dfrac{5}{8}+\dfrac{3}{8}\cos 4x$
Vậy $y=\dfrac{5}{4}+\dfrac{3}{4}\cos 4x+5=\dfrac{25}{4}+\dfrac{3}{4}\cos 4x$
có ngay $\dfrac{25}{4}-\dfrac{3}{4} \le y \le \dfrac{25}{4}+\dfrac{3}{4}$
Hay $\dfrac{11}{2}\le y \le 7$ dấu $=$ dễ nên tự tìm