Question

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

Answer 1

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