Đặt $t=\sin x+\cos x\Rightarrow -\sqrt 2\le t\le \sqrt 2$
$\begin{array}{l} {t^2} = {\left( {\sin x + \cos x} \right)^2} = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x\\ \Leftrightarrow {t^2} = 1 + 2\sin x\cos x \Rightarrow \sin x\cos x = \dfrac{{{t^2} - 1}}{2} \end{array}$
$\begin{array}{l} {\sin ^3}x + {\cos ^3}x - 1 = \dfrac{3}{2}\sin 2x\\ \Leftrightarrow {\sin ^3}x + {\cos ^3}x - 1 = \dfrac{3}{2}.2\sin x\cos x\\ \Leftrightarrow {\sin ^3}x + {\cos ^3}x - 1 = 3\sin x\cos x\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x - \sin x\cos x} \right) = 1 + 3\sin x\cos x\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 - \sin x\cos x} \right) = 1 + 3\sin x\cos x\\ \Leftrightarrow \left( {\sin x + \cos x} \right) - \sin x\cos x\left( {\sin x + \cos x} \right) = 1 + 3\sin x\cos x\\ \Leftrightarrow t - \dfrac{{t\left( {{t^2} - 1} \right)}}{2} = 1 + \dfrac{{3\left( {{t^2} - 1} \right)}}{2}\\ \Leftrightarrow 2t - t\left( {{t^2} - 1} \right) = 2 + 3\left( {{t^2} - 1} \right)\\ \Leftrightarrow 2t - {t^3} + t = 2 + 3{t^2} - 3\\ \Leftrightarrow {t^3} + 3{t^2} - 3t - 1 = 0\\ \Leftrightarrow \left( {t - 1} \right)\left( {{t^2} + 4t + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t = - 2 - \sqrt 3 \\ t = - 2 + \sqrt 3 \end{array} \right. \Rightarrow \left[ \begin{array}{l} t = 1\\ t = - 2 + \sqrt 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sin x + \cos x = 1\\ \sin x + \cos x = - 2 + \sqrt 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = - 2 + \sqrt 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\ x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \\ x + \dfrac{\pi }{4} = \dfrac{{\arcsin \left( { - 2 + \sqrt 3 } \right)}}{{\sqrt 2 }} + k2\pi \\ x + \dfrac{\pi }{4} = \pi - \dfrac{{\arcsin \left( { - 2 + \sqrt 3 } \right)}}{{\sqrt 2 }} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{2} + k2\pi \\ x = - \dfrac{\pi }{4} + \dfrac{{\arcsin \left( { - 2 + \sqrt 3 } \right)}}{{\sqrt 2 }} + k2\pi \\ x = \dfrac{{3\pi }}{4} - \dfrac{{\arcsin \left( { - 2 + \sqrt 3 } \right)}}{{\sqrt 2 }} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
