3. Find the largest integer \( a \) such that the expression

\[
a^{2}-15 a-(\tan x-1)(\tan x+2)(\tan x+5)(\tan x+8)
\]

is less than 35 for any value of \( x \in (-\pi / 2, \pi / 2) \).

(6 points)
Solution. Let's make the substitution $t=\operatorname{tg} x$. We need to determine for which values of $a$ the inequality $a^{2}-15 a-(t-1)(t+2)(t+5)(t+8)a^{2}-15 a-35,\left(t^{2}+7 t-8\right)\left(t^{2}+7 t+10\right)>a^{2}-15 a-35$

$z=t^{2}+7 t+1,(z-9)(z+9)>a^{2}-15 a-35, z^{2}>a^{2}-15 a+46$, $0>a^{2}-15 a+46, \sqrt{D}=\sqrt{41},(15-\sqrt{41}) / 2<a<(15-\sqrt{41}) / 2, \Rightarrow a=10$.

Answer: 10.