Đáp án: $x=k\pi$
Giải thích các bước giải:
Ta có:
$\sin^2x-\sin^4x+\cos^4x=1$
$\to\sin^2x+(\cos^4x-\sin^4x)=1$
$\to\sin^2x+(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)=1$
$\to\sin^2x+(\cos^2x-\sin^2x)\cdot1=1$
$\to\sin^2x+\cos^2x-\sin^2x=1$
$\to \cos^2x=1$
$\to \cos x=\pm1$
$\to x=k\pi$