1. Solve the inequality

$$
4^{\sin ^{2} \pi x}+3 \cdot 4^{\cos ^{2} \pi x} \leq 8
$$
Solution. If we denote $4^{\sin ^{2} \pi x}=y$, then

$$
4^{\cos ^{2} \pi x}=4^{1-\sin ^{2} \pi x}=\frac{4}{4^{\sin ^{2} \pi x}}=\frac{4}{y}
$$

the inequality takes the form $y+3 \cdot \frac{4}{y} \leq 8, y>0$. From this it follows that

$$
y^{2}-8 y+12 \leq 0, y>0
$$

from which we get

$$
2 \leq y \leq 6 \text {, i.e. } 2 \leq 4^{\sin ^{2} \pi x} \leq 6 \text {. }
$$

Since $0 \leq \sin ^{2} \pi x \leq 1$, it follows that

$$
1 \leq 4^{\sin ^{2} \pi x} \leq 4
$$

From (1) and (2) we get $2 \leq 4^{\sin ^{2} \pi x} \leq 4$, from which it follows that $\frac{1}{2} \leq \sin ^{2} \pi x \leq 1$ i.e $\frac{\sqrt{2}}{2} \leq|\sin \pi x| \leq 1$. The last inequality is equivalent to the inequalities $\sin \pi x \leq-\frac{\sqrt{2}}{2}$ or $\sin \pi x \geq \frac{\sqrt{2}}{2}$, from which $\frac{4 k+1}{4} \leq x \leq \frac{4 k+3}{4}, k \in \mathbb{Z}$.