For what $x\in[0,2\pi]$ is $\sin x < \cos 2x$ What's the set of all solutions to the inequality $\sin x < \cos 2x$ for $x \in [0, 2\pi]$? I know the answer is $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi]$, but I'm not quite sure how to get there.
This is what I have so far: $\\
\sin x - \cos 2x < 0\\
\sin x - \cos^2 x + \sin^2 x < 0\\
2\sin^2 x + \sin x - 1 < 0\\
-1 < \sin x < \frac{1}{2}$
Any help will be much appreciated.

You are doing fine, now just need to find the values of $x$ that make $\sin x$ come out to those values.  One way to see $\sin \frac \pi6=\frac 12$ is to consider an equilateral triangle cut in half vertically.  The base (adjacent) is $\frac 12$ and the hypotenuse is $1$.  To see $\sin \frac {3\pi}2=-1$ just consider the unit circle-the $\sin$ is $-1$ only straight down.
