Đáp án:
$\left[\begin{array}{l}x = \dfrac{1}{4}\arccos\dfrac{11}{13} + k\dfrac{\pi}{2}\\x = -\dfrac{1}{4}\arccos\dfrac{11}{13} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin^6x + \cos^6x = 4\cos^22x$
$\Leftrightarrow (\sin^2x + \cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x + \cos^2x) = 2(1 +\cos4x)$
$\Leftrightarrow 1 - \dfrac{3}{4}(2\sin x\cos x)^2 = 2 + 2\cos4x$
$\Leftrightarrow -\dfrac{3}{4}\sin^22x = 1 + 2\cos4x$
$\Leftrightarrow -\dfrac{3}{8}(1 - \cos4x) = 1 + 2\cos4x$
$\Leftrightarrow 13\cos4x = 11$
$\Leftrightarrow \cos4x = \dfrac{11}{13}$
$\Leftrightarrow \left[\begin{array}{l}4x = \arccos\dfrac{11}{13} + k2\pi\\4x = -\arccos\dfrac{11}{13} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{1}{4}\arccos\dfrac{11}{13} + k\dfrac{\pi}{2}\\x = -\dfrac{1}{4}\arccos\dfrac{11}{13} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)$
