Решить уравнение!cos2x=4корня из 5 sin(x-пи/2)-7

$cos(2x)=4\sqrt{5}sin(x-\frac{\pi}{2})-7$
$2cos^2x-1=-4\sqrt{5}cos x-7$
$cos x=t; -1 \leq t \leq 1$
$2t^2-1=-4\sqrt{5}t-7$
$2t^2+4\sqrt{5}t+6=0$
$t^2+2\sqrt{5}+3=0$
$D=(2\sqrt{5})^2-4*1*3=4*5-12=8=(2\sqrt{2})^2$
$t_1=\frac{-2\sqrt{5}-2\sqrt{2}}{2}=-\sqrt{5}-\sqrt{2}<-1$
$t_2=\frac{-2\sqrt{5}+2\sqrt{2}}{2}=\sqrt{2}-\sqrt{5}$
$cos x=\sqrt{2}-\sqrt{5}$
$x=^+_-arccos(\sqrt{2}-\sqrt{5})+2\pi*k$
k є Z