<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] $$\cos 2x = 2\sin x$$
or, $$1-2\sin^2 x = 2\sin x$$
or, $$2\sin^2 x + 2\sin x - 1 = 0$$
So $$\sin x = \frac{-2 \pm \sqrt{4-4\cdot (-1)\cdot 2}}{2\cdot 2}$$
$$=\frac{-2 \pm 2 \sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}=\omega, \omega^2$$
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