<s>[INST] Solving a sine inequality, need intuition We have the function $f(x) = \sin (2x - \dfrac{1}{3} \pi)$ on the domain $ [0, 1\dfrac{1}{2} \pi ]$. Solve the inequality: $f(x) > \dfrac{1}{2}$
So I got to this point (I wrote it as in equality first):
$$x= \dfrac{1}{4} \pi + k\pi \vee x = \dfrac{7}{12} \pi + k\pi$$
This would yield the solutions $\dfrac{1}{4}\pi$, $1\dfrac{1}{4}\pi$ and $\dfrac{7}{12}\pi$. But my problem is, I don't know when $f(x) > 0.5$. I need some intuition (preferably using the unit circle) to figure out what the conditions would be. 
 [/INST] $$\sin\left(2x-\frac{\pi}{3}\right)>\frac{1}{2}\;,\;\;x\in\left[0,\frac{3\pi}{2}\right]\iff \frac{\pi}{6}<2x-\frac{\pi}{3}<\frac{5\pi}{6}$$
Why? The easiest way to show the above is on the trigonometric (unit) circle: the sine function is positive only on the first and second quadrants, and it increases as the angle increases towards $\,\pi/2\,$  there...thus:
$$\frac{\pi}{6}<2x-\frac{\pi}{3}<\frac{5\pi}{6}\iff \frac{\pi}{2}<2x<\frac{7\pi}{6}\ldots$$
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