6.6. (New York, 73). Find all values of $x \in [0, \pi / 2]$, satisfying the equation $\cos ^{8} x + \sin ^{8} x = 97 / 128$.
6.6. Transforming the left side of the equation, we get

$$
\begin{aligned}
& \cos ^{8} x+\sin ^{8} x=\left(\frac{1+\cos 2 x}{2}\right)^{4}+\left(\frac{1-\cos 2 x}{2}\right)^{4}= \\
& =\frac{1}{16}\left(2+12 \cos ^{2} 2 x+2 \cos ^{4} 2 x\right)=\frac{1}{32}\left(4+12(1+\cos 4 x)+(1+\cos 4 x)^{2}\right)= \\
& =\frac{1}{32}\left(\cos ^{2} 4 x+14 \cos 4 x+17\right)
\end{aligned}
$$

Therefore, we have (an equation equivalent to the original one)

$$
\cos ^{2} 4 x+14 \cos 4 x=97 / 4-17=29 / 4
$$

i.e.,

$$
(\cos 4 x+29 / 2)(\cos 4 x-1 / 2)=0, \text { or } \cos 4 x=1 / 2
$$

Considering that $4 x \in[0 ; 2 \pi]$, we have $4 x \in\{\pi / 3 ; 5 \pi / 3\}$. Thus, we get the answer: $\pi / 12,5 \pi / 12$.