3. Determine all values of $x$ that belong to the interval $[0,2 \pi]$ for which the inequality

$$
2\left(\cos ^{2} x-\sqrt{3} \sin ^{2} x\right)>(\sqrt{3}-1) \sin ^{2} 2 x
$$

is true.
Solution. The given inequality is equivalent to the inequality

$$
2(\sqrt{3}-1) \sin ^{4} x+(1-3 \sqrt{3}) \sin ^{2} x+1>0
$$

Let $y_{1}, y_{2}\left(y_{1}<y_{2}\right)$ denote the solutions of the quadratic equation

$$
2(\sqrt{3}-1) y^{2}+(1-3 \sqrt{3}) y+1=0
$$

It is easily verified that for them $0<y_{1}<1<y_{2}$. Taking into account that $0 \leq \sin ^{2} x \leq 1$, we obtain that the given inequality is satisfied for $0 \leq \sin ^{2} x<y_{1}$, i.e., for $|\sin x|<\sqrt{y_{1}}$. Therefore, the values of $x \in[0,2 \pi]$ that satisfy the inequality are:

$$
0 \leq x<\arcsin \sqrt{y_{1}}, \pi-\arcsin \sqrt{y_{1}}<x<\pi+\arcsin \sqrt{y_{1}}, 2 \pi-\arcsin \sqrt{y_{1}}<x \leq 2 \pi
$$