Đáp án:
$x = \pm \dfrac{\pi}{3} + k\pi$ $(k \in \Bbb Z)$
Giải thích các bước giải:
$sin^6\dfrac{x}{2} + cos^6\dfrac{x}{2} = \dfrac{7}{16}$
$\Leftrightarrow \left(sin^2\dfrac{x}{2} + cos^2\dfrac{x}{2}\right)\cdot \left(sin^4\dfrac{x}{2} - sin^2\dfrac{x}{2}cos^2\dfrac{x}{2} + cos^4\dfrac{x}{2}\right) = \dfrac{7}{16}$
$\Leftrightarrow \left(sin^2\dfrac{x}{2}\right)^2 + 2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2} + \left(cos^2\dfrac{x}{2}\right)^2 - 3\left(sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2 = \dfrac{7}{16}$
$\Leftrightarrow 1 - \dfrac{3}{4}sin^2x = \dfrac{7}{16}$
$\Leftrightarrow 1 - \dfrac{3}{4}(1 - cos2x) = \dfrac{7}{16}$
$\Leftrightarrow 6cos2x = -3$
$\Leftrightarrow 2x = \pm \dfrac{2\pi}{3} + k2\pi$
$\Leftrightarrow x = \pm \dfrac{\pi}{3} + k\pi$ $(k \in \Bbb Z)$
