Đáp án:

$\text{D. M}= 1 - \dfrac{3}{4}\sin^2 2x$

Giải thích:

$M = \sin^6 x + \cos^6 x$

$= (\sin^2 x)^3 + (\cos^2 x)^3$

$= (\sin^2 x + \cos^2 x)[(\sin^2 x)^2 - \sin^2 x \cdot \cos^2 x + (\cos^2 x)^2]$

$= 1 \cdot [(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cdot \cos^2 x - \sin^2 x \cdot \cos^2 x]$

$= 1^2 - 3\sin^2 x \cdot \cos^2 x$

$= 1 - 3(\sin x \cdot \cos x)^2$

$= 1 - 3 \cdot \left(\dfrac{1}{2}\sin 2x\right)^2$

$= 1 - \dfrac{3}{4}\sin^2 2x$

$\Longrightarrow \text{ Đáp án đúng là D. M}= 1 - \dfrac{3}{4}\sin^2 2x$