Question

решите уровнение)
cos4x+2cos^2x=0

Answer 1

$cos 4x+2cos^2x=0\ (2cos^22x-1)+(1+cos 2x)=0\ 2cos^22x+cos 2x=0\ cos 2x(2cos 2x+1)=0\ left[ begin{matrix} cos 2x=0\ 2cos 2x+1=0 end{matrix}right <=> left[ begin{matrix} cos 2x=0\ cos 2x=-frac{1}{2} end{matrix}right <=> \ left[ begin{matrix} 2x=frac{pi}{2}+pi k\ 2x=pm frac{2pi}{3}+2pi m end{matrix}right <=> left[ begin{matrix} x=frac{pi}{4}+frac{pi k}{2}\ x=pm frac{pi}{3}+pi m end{matrix}right.$
$Ombem: frac{pi}{4}+frac{pi k}{2}; pm frac{pi}{3}+pi m; k in Z, m in Z.$