Question

Помогите с уравнением: 4^cos2x+4^cos^2 x=3

Answer 1

$4^{cos 2x}+4^{cos^2 x}=3$
$4^{2cos^2 x-1}+4^{cos^2 x}-3=0$
$frac{1}{4}*(4^{cos^2 x})^2+4^{cos^2 x}-3=0$
$4^{cos^2 }=t>0$
$frac{1}{4}t^2+t-3=0$
$t^2+4t-12=0$
$(t+6)(t-2)=0$
$t+6=0;t_1=-6<0$
$t-2=0;t_2=2$

$4^{cos^2 x}=2$
$2^{2cos^2 x}=2^1$
$2cos^2 x=1$
$cos^2 x=frac{1}{2}$
$frac{1+cos 2x}{2}=frac{1}{2}$
$1+cos 2x=1$
$cos 2x=0$
$2x=frac{pi}{2}+pi*k$
k є Z
$x=frac{pi}{4}+frac{pi*k}{2}$
k є Z