Giải thích các bước giải:
$\sin^4x+\cos^4x-\cos 2x+\dfrac 14\sin^22x+m=0$
$\to (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\cos 2x+\dfrac 14\sin^22x+m=0$
$\to 1-\dfrac 12\sin^22x-\cos 2x+\dfrac 14\sin^22x+m=0$
$\to 1-\dfrac 14\sin^22x-\cos 2x+m=0$
$\to 1-\dfrac 14(1-\cos^22x)-\cos 2x+m=0$
$\to \dfrac 34+\dfrac 14\cos^22x-\cos 2x+m=0$
$\to \dfrac 14\cos^22x-\cos 2x+\dfrac 34=-m$
$\to \cos^22x-4\cos 2x+3=-4m$
$\to (\cos 2x-2)^2=1-4m$
Mà $-1\le \cos 2x\le 1\to -3\le \cos 2x-2\le -1\to 0\le (\cos 2x-2)^2\le 3$
$\to 0\le 1-4m\le 3$
$\to \dfrac{-1}{2}\le m\le \dfrac 14$
