Given :
$4\sin ^4x+\cos^4x=1$
$\Rightarrow 4\sin^4x+(\cos^2x+1)(\cos^2x-1)=0$
$4\sin^4x-\sin^2x(1+\cos^2x)=0$
$\sin^2x(5\sin^2x-2)=0$
$\sin^2x=0$
Or $\sin^2x=\large\frac{2}{5}$
$x=n\pi$
$\sin^2x=\large\frac{2}{5}$
$\Rightarrow x=n\pi$
$x=n\pi$
Or $n\pi\pm\sin^{-1}\big(\sqrt{\large\frac{2}{5}}\big)$
Hence (d) is the correct answer.