Đáp án:
$x = \dfrac{\pi}{2}+k\pi\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\sin^4x -\cos^2x = 1$
$\Leftrightarrow (1 - \sin^4x) +\cos^2x = 0$
$\Leftrightarrow (1-\sin^2x)(1+\sin^2x) + \cos^2x = 0$
$\Leftrightarrow \cos^2x(1+\sin^2x) +\cos^2x = 0$
$\Leftrightarrow \cos^2x(\sin^2x + 2)=0$
$\Leftrightarrow \cos x = 0 \qquad (\sin^2x + 2> 0)$
$\Leftrightarrow x = \dfrac{\pi}{2}+k\pi\quad (k\in \Bbb Z)$