mark $\sin{x}=a$ and $\cos{x}=b$ the equation is equivalent to $a^3\left(1+\frac{b}{a}\right)+b^3\left(1+\frac{a}{b}\right)=b^2-a^2$ we consider two cases $a+b=0$ and $a+b\neq 0$ for $a+b \neq 0$ we get: $a^2+b^2=a-b \implies \sin2x=0$ clearly this don't works so we have $a+b=0$ which gives $\tan{x}=-1$ now this gives $x=\varphi\pi-\frac{\pi}{4} \qquad \forall \qquad \varphi \in \mathbb{Z}$

$\sin x+\cos x=\cos 2x\Rightarrow {{(\sin x+\cos x)}^{2}}={{\cos }^{2}}2x\Leftrightarrow 1+\sin 2x=1-{{\sin }^{2}}2x$thus $\sin 2x(1+\sin 2x)=0$