ĐK: $\cos x\ne 0\to x\ne \dfrac{\pi}{2}+k\pi$
$\tan^2x+1+\sin^22x=4\cos^2x+1$
$\to \dfrac{1}{\cos^2x}+1-\cos^22x-4\cos^2x-1=0$
$\to \dfrac{1}{\cos^2x}-(2\cos^2x-1)^2-4\cos^2x=0$
$\to \dfrac{1}{\cos^2x}-4\cos^4x+4\cos^2x-1-4\cos^2x=0$
$\to \dfrac{1}{\cos^2x}-4\cos^4x-1=0$
$\to 1-4\cos^6x-\cos^2x=0$
$\to \cos x=\pm\dfrac{\sqrt2}{2}$
$\to x=\pm\dfrac{\pi}{4}+k2\pi$ hoặc $x=\pm\dfrac{3\pi}{4}+k2\pi$
Kết hợp, ta có $x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
