Đáp án:
$\left[\begin{array}{l}x = - \dfrac{\pi}{4} + k\pi\\x =\dfrac{\pi}{2}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}1 + \sin^32x + \cos^32x = \dfrac{3}{2}\sin4x\\ \Leftrightarrow 1 + (\sin2x + \cos2x)^3 - 3\sin2x\cos2x(\sin2x + \cos2x) = 3\sin2x\cos2x \\ \Leftrightarrow (\sin2x + \cos2x)^3 - 3\sin2x\cos2x(\sin2x + \cos2x + 1) + 1 = 0\\ Đặt\,\,t = \sin2x + \cos2x \qquad (|t| \leq \sqrt2)\\ \Rightarrow t^2 = 1 + 2\sin2x\cos2x\\ \Rightarrow \dfrac{t^2 -1}{2} = \sin2x\cos2x\\ \text{Phương trình trở thành:}\\ t^3 - 3.\dfrac{t^2 - 1}{2}(t + 1) + 1 = 0\\ \Leftrightarrow -t^3 - 3t^2 + 3t + 5 = 0\\ \Leftrightarrow \left[\begin{array}{l}t \approx -3,44949 \quad (loại)\\t = -1\quad (nhận)\\t = 1,44949\quad (loại)\end{array}\right.\\ Với\,\,t = - 1\,\,ta\,\,được:\\ \sin2x + cos2x = - 1\\ \Leftrightarrow \sqrt2\sin\left(2x + \dfrac{\pi}{4}\right) = -1\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{4}\right) = - \dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{4} = - \dfrac{\pi}{4} + k2\pi\\2x + \dfrac{\pi}{4} = \dfrac{5\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{4} + k\pi\\x =\dfrac{\pi}{2}+ k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
