Task B-3.1. Solve the inequality

$$
\frac{\sin ^{2} x-2 \sin 2 x+3 \cos ^{2} x}{1-\sin 2 x} \geq 0, \quad \text { if }|x|<\frac{\pi}{2}
$$
Solution. Since $|\sin 2 x| \leq 1, \forall x \in \mathbb{R}$, it follows that $1-\sin 2 x \geq 0, \forall x \in \mathbb{R}$, hence the given inequality is equivalent to the inequality

$$
\sin ^{2} x-2 \sin 2 x+3 \cos ^{2} x \geq 0, \quad \text { with the condition } \sin 2 x \neq 1, \text { i.e., } x \neq \frac{\pi}{4}
$$

By applying the formula for $\sin 2 x$, we get

$$
\sin ^{2} x-4 \sin x \cos x+3 \cos ^{2} x \geq 0
$$

Since $x \neq \pm \frac{\pi}{2}$, it follows that $\cos x \neq 0$ and we can divide the given inequality by $\cos ^{2} x$. This yields

$$
\operatorname{tg}^{2} x-4 \operatorname{tg} x+3 \geq 0
$$

Let's introduce the substitution $\operatorname{tg} x=t$.

We obtain the quadratic inequality $t^{2}-4 t+3 \geq 0$ whose solutions are $t \leq 1$ or $t \geq 3$. This implies $\operatorname{tg} x \leq 1$ or $\operatorname{tg} x \geq 3$. We seek the solutions of this inequality only on the interval $\left\langle-\frac{\pi}{2}, \frac{\pi}{2}\right\rangle$.

![](https://cdn.mathpix.com/cropped/2024_05_30_2595713ee0b414c2653fg-11.jpg?height=718&width=506&top_left_y=1652&top_left_x=777)

From the graph, we read the solution, with the condition $x \neq \frac{\pi}{4}$,

$$
x \in\left\langle-\frac{\pi}{2}, \frac{\pi}{4}\right\rangle \cup\left[\operatorname{arctg} 3, \frac{\pi}{2}\right\rangle
$$