B1. Find all real numbers $x$ that satisfy the inequality

$$
2 \sin ^{2} 2 x \geq 3 \cos 2 x
$$
B1. Using the formulas for double-angle trigonometric functions, we get

$$
8 \sin ^{2} x \cos ^{2} x \geq 3 \cos ^{2} x-3 \sin ^{2} x
$$

By considering the relationship $\cos ^{2} x=1-\sin ^{2} x$, we rearrange the inequality to

$$
8 \sin ^{4} x-14 \sin ^{2} x+3 \leq 0
$$

If we introduce a new variable $y=\sin ^{2} x$, we get the inequality $8 y^{2}-14 y+3 \leq 0$ or $(4 y-1)(2 y-3) \leq 0$. Since $0 \leq y \leq 1$, the term $(2 y-3)$ is automatically negative. Therefore, the given inequality is equivalent to the inequality $4 y-1 \geq 0$ or $\sin ^{2} x \geq \frac{1}{4}$. The latter inequality holds when either $\sin x \geq \frac{1}{2}$, i.e., $\frac{\pi}{6}+2 k \pi \leq x \leq \frac{5 \pi}{6}+2 k \pi, k \in \mathbb{Z}$, or $\sin x \leq-\frac{1}{2}$, i.e., $-\frac{5 \pi}{6}+2 k \pi \leq x \leq-\frac{\pi}{6}+2 k \pi, k \in \mathbb{Z}$. The combined solutions are thus all numbers $x$ that satisfy the inequality $\frac{\pi}{6}+k \pi \leq x \leq \frac{5 \pi}{6}+k \pi$ for some integer $k$.
Writing the equivalent inequality in single angles ..... 1 point
Writing the inequality $8 \sin ^{4} x-14 \sin ^{2} x+3 \leq 0$ ..... 1 point
Conclusion that the factor $2 \sin ^{2} x-3$ is negative ..... 1 point
Writing the equivalent inequalities $\sin x \geq \frac{1}{2}$ or $\sin x \leq-\frac{1}{2}$ ..... 1 point
Writing the solutions ..... 2 points