Question

Решите уравнение cos^2 x+cos^2 2x+cos^ 3x+cos^2 4x=25 задание

Answer 1

$cos^2x+cos^22x+cos^23x+cos^24x=2$

$2cos^2x+2cos^22x+2cos^23x+2cos^24x=4$

$2cos^2x-1+2cos^22x-1+2cos^23x-1+2cos^24x-1=0$

$cos2x+cos4x+cos6x+cos8x=0$

Далее пользуемся формулой:  $cosa-cosb=2*cos(\frac{a+b}2)*cos(\frac{a-b}2)$

$2*cos(\frac{2x+8x}2)*cos(\frac{2x-8x}2)+2*cos(\frac{4x+6x}2)*cos(\frac{4x6x}2)=0$

$cos(5x)*cos(3x)+cos(5x)*cos(x)=0$

$cos(5x)(cos(3x)+cos(x))=0$

$cos(5x)*2*cos(\frac{3x+x}2)*cos(\frac{3x-x}2)=0$

$cos(5x)*cos(2x)*cos(x)=0$

$cos(5x)=0$

$5x=\frac{\pi}2+\pi n;n\in Z$

$x=\frac{\pi}{10}+\frac{\pi n}5;n\in Z$

$cos(2x)=0$

$2x=\frac{\pi}{2}+\pi n;n\in Z$

$x=\frac{\pi}{4}+\frac{\pi n}2;n\in Z$

$cos(x)=0$

$x=\frac{\pi}{2}+\pi n;n\in Z$