TXĐ: $D=R$
$2cos^22x-sin^2x-2=0$
$↔ 2(1-2sin^2x)^2-sin^2x-2=0$
Đặt $sin^2x=t$, $0≤t≤1$, ta có:
$2(1-2t)^2-t-2=0$
$↔ 2(4t^2-4t+1)-t-2=0$
$↔ 8t^2-9t=0$
$↔ t(8t-9)=0$
$↔ \left[ \begin{array}{l}t=0\\t=\dfrac{9}{8}\end{array} \right.$
Loại $t=\dfrac{9}{8}$ vì $0≤t≤1$
Ta có: $sin^2x=0$
$↔ sinx=0$
$↔ x=k\pi$ $(k∈Z)$ (thỏa mãn)