<s>[INST] How would you solve for x in this case using the trig identities? 
Solve for all values of $x$:
$$\cos 2x = 2\sin x$$
$$1 - 2\sin^2x - 2\sin x = 0$$
$$-2\sin^2x - 2\sin x + 1 = 0$$
$$\sin^2x - 2\sin x - 2 = 0$$

How would you factor this above to solve for $x$?
 [/INST] First there needs to be a little algebra and trigonometry,
$$\cos(2x)=2\sin(x)$$
$$1-2\sin^2(x)=2\sin(x)$$
$$1=2\sin^2(x)-2\sin(x)$$
$$2\sin^2(x)-2\sin(x)-1=0$$
$$\sin x = \frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}$$
Let $z=\sin(x)$. Then,
$$2\sin^2(x)-2\sin(x)-1=0\implies2z^2-2z-1=0$$
Now use quadratic formula,
$$z=\frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}=z$$ 
Since $z=\sin(x)$, then
$$x=\sin^{-1}(z)$$
$$x=\sin^{-1}(\frac{-1\pm\sqrt{3}}{2})$$
$$x=\sin^{-1}(\frac{-1+\sqrt{3}}{2}), \ x=\sin^{-1}(\frac{-1-\sqrt{3}}{2})$$
The rest is trivial: just calculate!
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