Đáp án: $ x=\dfrac12k\pi, k\in Z$
Giải thích các bước giải:
Ta có:
$\cos^4x+\sin^4x=-\sin4x$
$\to (\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x=-\sin4x$
$\to 1^2-\dfrac12\cdot (2\sin x\cos x)^2=-\sin4x$
$\to 1-\dfrac12\cdot \sin^22x=-(1-2\sin^22x)$
$\to 1-\dfrac12\cdot \sin^22x=-1+2\sin^22x$
$\to \dfrac52\sin^22x=0$
$\to\sin^22x=0$
$\to\sin2x=0$
$\to 2x=k\pi, k\in Z$
$\to x=\dfrac12k\pi$
