8. The function $y=f(t)$ is such that the sum of the roots of the equation $f(\sin x)=0$ on the interval $[3 \pi / 2, 2 \pi]$ is $33 \pi$, and the sum of the roots of the equation $f(\cos x)=0$ on the interval $[\pi, 3 \pi / 2]$ is $23 \pi$. What is the sum of the roots of the second equation on the interval $[\pi / 2, \pi]$?
Answer: $17 \pi$.

Solution. Let the equation $f(\sin x)=0$ on the interval $\left[\frac{3 \pi}{2}, 2 \pi\right]$ have $k$ roots $x_{1}, x_{2}, \ldots, x_{k}$. This means that the equation $f(t)=0$ on the interval $[-1 ; 0]$ has $k$ roots $t_{1}, \ldots, t_{k}$. Since $\arcsin t_{i} \in$ $\left[-\frac{\pi}{2}, 0\right]$, then $x_{i}=2 \pi+\arcsin t_{i}$ and

$$
2 \pi k+\arcsin t_{1}+\ldots+\arcsin t_{k}=33 \pi
$$

Similarly, the equation $f(\cos x)=0$ on the interval $\left[\pi, \frac{3 \pi}{2}\right]$ also has $k$ roots, which can be written as $2 \pi-\arccos t_{i}$ (since $\arccos t_{i} \in\left[\frac{\pi}{2}, \pi\right]$), so

$$
2 \pi k-\arccos t_{1}-\ldots-\arccos t_{k}=23 \pi
$$

From the two obtained equations, it follows that

$$
\arcsin t_{1}+\ldots+\arcsin t_{k}+\arccos t_{1}+\ldots+\arccos t_{k}=10 \pi
$$

Since $\arcsin \alpha+\arccos \alpha=\frac{\pi}{2}$, then $\frac{\pi}{2} k=10 \pi$, from which $k=20$.

Finally, on the interval $\left[\frac{\pi}{2}, \pi\right]$ the equation $f(\cos x)=0$ also has $k=20$ roots $\arccos t_{1}, \ldots$, $\arccos t_{20}$, for which we have

$$
\arccos t_{1}+\ldots+\arccos t_{20}=2 \cdot 20 \pi-23 \pi=17 \pi
$$

## Variant 2

Answers. 1. 11; 2. $3 ; \mathbf{3} .8 \sqrt{2} ; \mathbf{4} .\left(\frac{-2-\sqrt{5}}{3} ;-\frac{4}{3}\right) \cup\left(\frac{-2+\sqrt{5}}{3} ; \infty\right) ; \mathbf{5 .} A K=8 ; \mathbf{6} . a=-6, b=4$, $c=1 ; \mathbf{7} .12 ; \mathbf{8} .43 \pi$.

## Variant 3

Answers. 1. $15 ; \mathbf{2} .2 ; \mathbf{3} .5 \sqrt{6} ; \mathbf{4} \cdot\left(\frac{-1-\sqrt{5}}{14} ;-\frac{1}{7}\right) \cup\left(\frac{-1+\sqrt{5}}{14} ; \infty\right) ;$ 5. $B L=9 ; \mathbf{6} . a=-1, b=4, c=6$; 7. $11 ; 8.19 \pi$.