9.1. Find the number of roots of the equation

$$
\operatorname{arctg}\left(\operatorname{tg}\left(\sqrt{13 \pi^{2}+12 \pi x-12 x^{2}}\right)\right)=\arcsin \left(\sin \sqrt{\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}}\right)
$$
Answer: 9. Solution. Let's make the substitution $t=\sqrt{\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}}$. Since $\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}=4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}$, then $0 \leq t \leq 2 \pi$. The original equation, after the specified substitution, transforms into the equation $\operatorname{arctg}(\operatorname{tg} 2 t)=\arcsin (\sin t)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2cb18ec48844428a3b16g-13.jpg?height=563&width=1171&top_left_y=201&top_left_x=497)

The graphs of the functions $f(t)=\operatorname{arctg}(\operatorname{tg} 2 t)$ and $g(t)=\arcsin (\sin t)$ on the interval $[0 ; 2 \pi]$ have five points of intersection: $t_{1}=0, t_{2} \in\left(\frac{\pi}{2} ; \pi\right), t_{3}=\pi, t_{4} \in\left(\pi ; \frac{3 \pi}{2}\right)$, and $t_{5}=2 \pi$.

The value of $t_{2}$ is found from the system: $\left\{\begin{array}{c}y=\pi-t, \\ y=2\left(t-\frac{\pi}{2}\right)\end{array} \Leftrightarrow t_{2}=\frac{2 \pi}{3}\right.$.

Similarly, we find $t_{4}:\left\{\begin{array}{c}y=\pi-t, \\ y=2\left(t-\frac{3 \pi}{2}\right)\end{array} \Leftrightarrow t_{4}=\frac{4 \pi}{3}\right.$.

The equations $\sqrt{4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}}=t_{k}$ have two distinct solutions for $k=1,2,3,4$. The equation $\sqrt{4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}}=t_{6}$ has a unique solution. In total, we get 9 solutions.