4. Solve the system of equations $\left\{\begin{array}{l}2 \cos ^{2} x+2 \sqrt{2} \cos x \cos ^{2} 4 x+\cos ^{2} 4 x=0, \\ \sin x=\cos y .\end{array}\right.$.

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# Solution:

Solve the 1st equation of the system:

$$
\begin{aligned}
& 2 \cos ^{2} x+2 \sqrt{2} \cos x \cos ^{2} 4 x+\cos ^{2} 4 x=0 \quad \Leftrightarrow\left(\sqrt{2} \cos x+\cos ^{2} 4 x\right)^{2}+\cos ^{2} 4 x \sin ^{2} 4 x=0 \\
& \left\{\begin{array} { l } 
{ \sqrt { 2 } \cos x + \cos ^ { 2 } 4 x = 0 , } \\
{ \cos ^ { 2 } 4 x \sin ^ { 2 } 4 x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \cos x = 0 , } \\
{ \cos 4 x = 0 , }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
\sqrt{2} \cos x+1=0, \\
\sin 4 x=0 .
\end{array} \quad\right.\right.\right. \text { The first system }
\end{aligned}
$$

has no solutions. The second system has solutions $x= \pm \frac{3 \pi}{4}+2 \pi k, k \in Z$.

Returning to the original system of equations, we obtain an equivalent system

$$
\left\{\begin{array} { l } 
{ x = \pm \frac { 3 \pi } { 4 } + 2 \pi k , k \in Z , } \\
{ \sin x = \cos y . }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ \{ \begin{array} { l } 
{ x = \frac { 3 \pi } { 4 } + 2 \pi k , k \in Z , } \\
{ \cos y = \sqrt { 2 } / 2 ; }
\end{array} } \\
{ \{ \begin{array} { l } 
{ x = - \frac { 3 \pi } { 4 } + 2 \pi k , k \in Z , } \\
{ \cos y = - \sqrt { 2 } / 2 }
\end{array} }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
x=\frac{3 \pi}{4}+2 \pi k, k \in Z \\
y= \pm \frac{\pi}{4}+2 \pi n, n \in Z
\end{array}\right. \\
\left\{\begin{array}{l}
x=-\frac{3 \pi}{4}+2 \pi k, k \in Z \\
y= \pm \frac{3 \pi}{4}+2 \pi n, n \in Z
\end{array}\right.
\end{array}\right.\right.\right.
$$

Answer: $\left(\frac{3 \pi}{4}+2 \pi k, \pm \frac{\pi}{4}+2 \pi n\right),\left(-\frac{3 \pi}{4}+2 \pi k, \pm \frac{3 \pi}{4}+2 \pi n\right), \quad k, n \in Z$.