2. Zadatak

Primijenimo li osnovni trigonometrijski identitet $\sin^2\alpha+\cos^2\alpha=1$ dobivamo

$$\begin{aligned} 4^{\sin^2\pi x}+3\cdot 4^{\cos^2\pi x}&\leq 8\\ \Longleftrightarrow\quad 4^{\sin^2\pi x}+3\cdot 4^{1-\sin^2\pi x}&\leq 8\\ \Longleftrightarrow\quad 4^{\sin^2\pi x}+12\cdot 4^{-\sin^2\pi x}&\leq 8. \end{aligned}$$

Uvođenjem supstitucije $t=4^{\sin^2\pi x}$ nejednadžba prelazi u

$$t+\frac{12}{t}\leq 8,$$ $$t^2-8t+12\leq 0.$$

Rješenje ove nejednadžbe je $t\in[2,6]$

pa dalje dobivamo

$$2\leq 4^{\sin^2\pi x}\leq 6,$$ $$\frac12\leq\sin^2\pi x\leq\log_4 6,$$ $$\frac{1}{\sqrt2}\leq|\sin\pi x|\leq\sqrt{\log_4 6}.$$

Kako je $\log_4 6>1$, nejednakost $|\sin\pi x|\leq\sqrt{\log_4 6}$ vrijedi za sve $x\in\mathbb{R}$.

Preostaje riješiti $|\sin\pi x|\geq\frac{1}{\sqrt2}$. To je ekvivalentno s

$$\sin\pi x\geq\frac{\sqrt2}{2} \quad\text{ili}\quad \sin\pi x\leq-\frac{\sqrt2}{2}.$$

Dakle, uz $k\in\mathbb{Z}$,

$$\frac{\pi}{4}+2k\pi\leq\pi x\leq\frac{3\pi}{4}+2k\pi \quad\text{ili}\quad -\frac{3\pi}{4}+2k\pi\leq\pi x\leq-\frac{\pi}{4}+2k\pi,$$

tj.

$$\frac14+2k\leq x\leq\frac34+2k \quad\text{ili}\quad -\frac34+2k\leq x\leq-\frac14+2k.$$

Kraće možemo zapisati

$$\frac14+k\leq x\leq\frac34+k$$

te konačno

$$x\in\bigcup_{k\in\mathbb{Z}}\left[\frac14+k,\frac34+k\right].$$