Áp dụng hằng đẳng thức ta có
$\cos^6x + \sin^6x - 3(\sin^4x + \cos^4x) = -2$
$<-> (\cos^2x + \sin^2x)(\cos^4x + \sin^4x - \sin^2x \cos^2x) - 3(\sin^4x + \cos^4x) = -2$
$<-> -2(\sin^4x + \cos^4x) - \sin^2x \cos^2x = -2$
$<-> 2[(\sin^2x + \cos^2x)^2 - 2\sin^2x \cos^2x] + \sin^2x \cos^2x = 2$
$<-> 2 - 4\sin^2x \cos^2x + \sin^2x \cos^2x = 2$
$<-> \sin^2x \cos^2x = 0$
$<-> \sin(2x) = 0$
Vậy $2x = k\pi$ hay $x = \dfrac{k\pi}{2}$
Vậy $x = \dfrac{k\pi}{2}, k \in \mathbb{Z}$.
