Cos 2x=1+4cos x Решить уравнение

$cos2x=1+4cosx$
$cos^{2}x-sin^{2}x=1+4cosx$
$cos^{2}x-1+cos^2x=1+4cosx$
$2cos^{2}x-4cosx-2=0$
$cos^{2}x-2cosx-1=0$
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