Factor the first two terms:
$$2\sin^2(x)\left(2\sin(2x)-3) \right)\left(\sin^2(x)-1 \right)-1=0$$
Apply a Pythagorean identity to the second factor of the first term and multiply through by $-2$:
$$4\sin^2(x)\cos^2(x)\left(2\sin(2x)-3) \right)+2=0$$
To the first three factors of the first term, apply the double-angle identity for sine:
$$\sin^2(2x)\left(2\sin(2x)-3) \right)+2=0$$
Distribute to obtain a cubic in $\sin(2x)$:
$$2\sin^3(2x)-3\sin^2(2x)+2=0$$
Factor:
$$\left(\sin(2x)-1 \right)\left(\sin^2(2x)-2\sin(2x)-2 \right)=0$$
Apply the zero factor property:
i) The first factor implies
$$\sin(2x)=1$$
$$x=\frac{\pi}{4}(4k+1)$$ where $$k\in\mathbb{Z}$$
ii) The second factor implies, by applying the quadratic formula:
$$\sin(2x)=1\pm\sqrt{3}$$
Discarding the root whose magnitude is greater than unity, we are left with:
$$\sin(2x)=1-\sqrt{3}$$
$$x=k\pi-\frac{1}{2}\sin^{-1}\left(\sqrt{3}-1 \right)$$
Using the identity $\sin(\pi-\theta)=\sin(\theta)$ we also have:
$$x=\frac{\pi}{2}(2k+1)+\frac{1}{2}\sin^{-1}\left(\sqrt{3}-1 \right)$$