Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{16}+ k\dfrac{\pi}{4}\\x = -\dfrac{\pi}{8}+ k\dfrac{\pi}{2}\end{array}\right.\quad(k\in\Bbb Z)$
Giải thích các bước giải:
$(\sin x + \cos x)^2 = 2\cos^3x$
$\Leftrightarrow \sin^2 + 2\sin x\cos x +\cos^2x = 2\dfrac{1 + \cos6x}{2}$
$\Leftrightarrow 1 + \sin2x = 1 + \cos6x$
$\Leftrightarrow \sin2x = \cos6x$
$\Leftrightarrow \sin2x = \sin\left(\dfrac{\pi}{2} - 6x\right)$
$\Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{2} - 6x + k2\pi\\2x = \dfrac{\pi}{2} + 6x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{16}+ k\dfrac{\pi}{4}\\x = -\dfrac{\pi}{8}+ k\dfrac{\pi}{2}\end{array}\right.\quad(k\in\Bbb Z)$
