Why is the double angle formula not used in this solution? solve $\sin 2x = 0.5$
I am looking at the answer to this question and it does not use the double angle formula to make this:
$\sin 2x = 2\sin x \cos x$
And instead use
$2x = \arcsin 0.5$
Why is the double angle formula not used in this situation?

Using the double angle formula here ends you up with two things that must multiply together to become $0.5$, and we would need to solve a nasty quadratic to discover $\sin x$ and $\cos x$.  Just using $\arcsin$ directly makes your life easier: $\arcsin 0.5$ is very easy to obtain, and we reduce the rest of the problem to simple manipulation.

To give some sense here, using double angle formula you get
$$2\sin x \cos x = \frac{1}{2}$$
This with the pythagorean identity $\sin^2 x + \cos^2 x = 1$ can combine to make the following two equations:
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{3}{2}$$
$$\sin^2 x - 2\sin x \cos x + \cos^2 x = \frac{1}{2}$$
Factor the left side:
$$(\sin x + \cos x)^2 = \frac{3}{2}$$
$$(\sin x - \cos x)^2 = \frac{1}{2}$$
Take the square root:
$$\sin x + \cos x = \pm\sqrt{\frac{3}{2}}$$
$$\sin x - \cos x = \pm\sqrt{\frac{1}{2}}$$
Add the two equations:
$$2\sin x = \pm\sqrt{\frac{3}{2}} \pm \sqrt{\frac{1}{2}}$$
And now finally we can discover the sine of the original angle:
$$\sin x = \pm \frac{\sqrt{6}\pm\sqrt{2}}{4}$$
And then we need to apply $\arcsin$ and do some fiddling with "which equivalent angles do I take?".  What a mess.  Instead, just find $\arcsin \frac{1}{2} = 2x$ and use supplemental and coterminal angles to find the other values of $2x$ that work.
