1. For what values of $x$ does the expression $\cos ^{2}(\pi \cos x)+\sin ^{2}(2 \pi \sqrt{3} \sin x)$ take its minimum value
1. Answer: $x= \pm \frac{\pi}{3}+\pi t, t \in Z$. Solution. Let's check if the sine and cosine in the given expression can simultaneously be zero for some values of $x$. If so, the expression takes its minimum value at these values of $x$.

$$
\left\{\begin{array} { c } 
{ \operatorname { c o s } ( \pi \operatorname { c o s } x ) = 0 } \\
{ \operatorname { s i n } ( 2 \pi \sqrt { 3 } \operatorname { s i n } x ) = 0 }
\end{array} \rightarrow \left\{\begin{array}{c}
\cos x=1 / 2+k, k \in Z \\
\sin x=\frac{m}{2 \sqrt{3}}, m \in Z
\end{array}\right.\right.
$$

Considering the range of cosine, there are two possible values for $k=0, k=-1$. Case 1. $k=0$

$$
\cos x=1 / 2 \rightarrow \cos ^{2} x=\frac{1}{4} \rightarrow \sin ^{2} x=\frac{3}{4}=\frac{m^{2}}{12} \rightarrow m^{2}=9 \rightarrow\left[\begin{array}{c}
m=3 \\
m=-3
\end{array}\right.
$$

1) $\left\{\begin{array}{c}\cos x=\frac{1}{2} \\ \sin x= \pm \frac{\sqrt{3}}{2}\end{array} \rightarrow x= \pm \frac{\pi}{3}+2 \pi t, t \in Z\right.$

Case 2. $k=-1$

$$
\cos x=-1 / 2 \rightarrow \cos ^{2} x=\frac{1}{4} \rightarrow \sin ^{2} x=\frac{3}{4}=\frac{m^{2}}{12} \rightarrow m^{2}=9 \rightarrow\left[\begin{array}{c}
m=3 \\
m=-3
\end{array}\right.
$$

2) $\left\{\begin{array}{l}\cos x=-\frac{1}{2} \\ \sin x= \pm \frac{\sqrt{3}}{2}\end{array} \rightarrow x= \pm \frac{2 \pi}{3}+2 \pi t, t \in Z\right.$

Combining the series, we get the answer $x= \pm \frac{\pi}{3}+\pi t, t \in Z$