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

$4^{cos 2x}+4^{cos^2 x}=3$
$4^{2cos^2 x-1}+4^{cos^2 x}-3=0$
$\frac{1}{4}*(4^{cos^2 x})^2+4^{cos^2 x}-3=0$

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

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