Question

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Найдите среднее арифметическое корней уравнения ctg^2(πx/2) - (1+√2)/sin(πx/2) + √2 + 1 = 0 на отрезке [-5;1].

Answer 1

$ctg^2 \, (\frac{\pi x}{2})-\frac{(1+\sqrt{2})}{\sin{(\frac{\pi x}{2})}}+\sqrt{2}+1=0 \\ \\ \sin^2{(\frac{\pi x}{2})}\neq 0; \ \ \ \sin{(\frac{\pi x}{2})}\neq 0 \\ \\ \frac{\pi x}{2}\neq \pi n, \ n\in Z \\ \\ x\neq 2n, \ n\in Z$

$ctg^2 \, (\frac{\pi x}{2})-\frac{(1+\sqrt{2})}{\sin{(\frac{\pi x}{2})}}+\sqrt{2}+1=0 \ \ \ \cdot | \sin^2{(\frac{\pi x}{2})} \\ \\ \cos^2{(\frac{\pi x}{2})}-(1+\sqrt{2})\cdot \sin{(\frac{\pi x}{2})}+(\sqrt{2}+1)\cdot \sin^2{(\frac{\pi x}{2})}=0 \\ \\ 1-\sin^2{(\frac{\pi x}{2})}-(1+\sqrt{2})\cdot \sin{(\frac{\pi x}{2})}+(\sqrt{2}+1)\cdot \sin^2{(\frac{\pi x}{2})}=0 \\ \\ t=\sin{(\frac{\pi x}{2})} \\ \\ 1-t^2-(1+\sqrt{2})\cdot t+(\sqrt{2}+1)\cdot t^2=0$

$(-t^2 +(1+\sqrt{2})\cdot t^2)-(1+\sqrt{2})\cdot t+1=0 \\ \\ t^2\cdot (-1+1+\sqrt{2})-(1+\sqrt{2})\cdot t +1=0 \\ \\ \sqrt{2} \cdot t^2-(1+\sqrt{2})\cdot t +1=0 \\ \\ \sqrt{2} \cdot t^2 -t-\sqrt{2}\cdot t+1=0 \\ \\ \sqrt{2}\cdot t^2+1 -\sqrt{2}\cdot t -t=0 \\ \\ (\sqrt{2}\cdot t^2-t)+(1-\sqrt{2}\cdot t) =0 \\ \\ -t\cdot (1-\sqrt{2}\cdot t)+(1-\sqrt{2}\cdot t) =0 \\ \\ (1-\sqrt{2}\cdot t)\cdot (1-t)=0$

1)

$1-\sqrt{2}\cdot t =0 \\ \\ \sqrt{2}\cdot t=1 \\ \\ t=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

$\sin{({\frac{\pi x}{2}})}=\frac{\sqrt{2}}{2} \\ \\ \frac{\pi x_1}{2}=\frac{\pi}{4}+2\pi n, \ n\in Z; \ \ \ \ \frac{\pi x_2}{2}=\frac{3\pi}{4}+2\pi n, \ n\in Z \\ \\ x_1=\frac{1}{2}+4n, \ n\in Z; \ \ \ \ \ \ \ \ x_2=\frac{3}{2}+4n, \ n\in Z$

2)

$1-t=0 \\ \\ t=1$

$\sin{(\frac{\pi x}{2})}=1 \\ \\ \frac{\pi x_3}{2}=\frac{\pi}{2}+2\pi n, \ n \in Z \\ \\ x_3=1+4n, \ n\in Z$

$[-5;1]$

$x_1=\frac{1}{2}+4\cdot (-1)=\frac{1}{2}-4=\frac{1-8}{2}=-\frac{7}{2}=-3,5$

$x_2=\frac{1}{2}+4\cdot 0=\frac{1}{2}=0,5$

$x_3=\frac{3}{2}+4\cdot (-1) = \frac{3}{2}-4=\frac{3-8}{2}=-\frac{5}{2}=-2,5$

$x_4=1+4\cdot (-1) =1-4=-3 \\ \\ x_5=1+4\cdot 0 =1$

$\frac{-3,5+0,5+(-2,5)+(-3)+1}{5}=\frac{-3-2,5-3+1}{5}=\frac{-7,5}{5}=-\frac{75}{50}=-\frac{3}{2}=-1,5$