Question

Cos 2x=1+4cos x
Решить уравнение

Answer 1

$cos2x=1+4cosx$
$cos^{2}x-sin^{2}x=1+4cosx$
$cos^{2}x-1+cos^2x=1+4cosx$
$2cos^{2}x-4cosx-2=0$
$cos^{2}x-2cosx-1=0$
Заменяем косинус
$cosx=y$
$y^{2}-2y-1=0$
$y=frac{1+- sqrt{2}}{2}$
$cosx=frac{1+ sqrt{2}}{2}$
$x=arccos(frac{1+ sqrt{2}}{2})+2 pi k$, k∈Z
$cosx=frac{1- sqrt{2}}{2}$
$x=arccos(frac{1- sqrt{2}}{2})+2 pi k$, k∈Z