$\sin^6x + \cos^6x = \dfrac{1}{4}\sin^2x$
$\Leftrightarrow (\sin^2x + \cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x + \cos^2x) = \dfrac{1}{4}\sin^2x$
$\Leftrightarrow 1 - 3\sin^2x\cos^2x = \dfrac{1}{4}\sin^2x$
$\Leftrightarrow 4 - 12\sin^2x(1 -\sin^2x) = \sin^2x$
$\Leftrightarrow 12\sin^4x - 13\sin^2x + 4 = 0$ (vô nghiệm)
Phương trình đã cho vô nghiệm