<s>[INST] 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.
 [/INST] You reached the right inequality. A slightly different way of finishing goes as follows.
We are interested in where $(2\sin x-1)(\sin x+1)\lt 0$.
We find the places where $(2\sin x-1)(\sin x+1)=0$. This happens at $x=\frac{\pi}{6}$, $x=\frac{5\pi}{6}$, and $x=\frac{3\pi}{2}$. 
These $3$ points divide the interval $[0,2\pi]$ into $4$ pieces (intervals). On each of these pieces, our function does not change sign. So in each piece, we can choose a convenient test point to find the sign of our function on the whole piece.
For example, on the piece $[0,\frac{\pi}{6})$ choose the test point $x=0$. Result: our function is $\lt 0$ at the test point, and therefore $\lt 0$ on the whole interval $[0,\frac{\pi}{6})$.
On the piece $(\frac{\pi}{6}, \frac{5\pi}{6})$, choose the test point $x=\frac{\pi}{2}$. There our function is positive. So we do not use this interval.
Two more pieces to go.
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