3.394. Given the function $f(x)=\cos ^{4} x+\sin ^{4} x$. Find $f(\alpha)$, if it is known that $\sin 2 \alpha=\frac{2}{3}$.
## Solution.

$$
\begin{aligned}
& f(x)=\cos ^{4} x+\sin ^{4} x=\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-2 \cos ^{2} x \sin ^{2} x= \\
& =1-2 \cos ^{2} x \sin ^{2} x=1-\frac{1}{2}\left(4 \cos ^{2} x \sin ^{2} x\right)=1-\frac{1}{2} \sin ^{2} 2 x \\
& f(\alpha)=1-\frac{1}{2} \sin ^{2} 2 \alpha=1-\frac{1}{2}(\sin 2 \alpha)^{2}=1-\frac{1}{2}\left(\frac{2}{3}\right)^{2}= \\
& =1-\frac{1}{2} \cdot \frac{4}{9}=1-\frac{2}{9}=\frac{7}{9}
\end{aligned}
$$

Answer: $f(\alpha)=\frac{7}{9}$.