Trigonometric Inequality I have a problem about this trigonometric inequality, which I cannot completely solve. In particular, I cannot get the whole solution the book provides and what a bad luck: I don't have the book with me, because this problem arose from one of my student's problem during a private lesson. 
$$\sin\left(\frac{x}{2}\right)\left(2\cos(x) - \sqrt{3}\right) <0$$
We have to solve it by invoking the unitary circle method, and all the related blabla questions. 
The fact is that i get those two solutions (before unifying them)
$$\pi + 2k\pi < \frac{x}{2} < 2\pi + 2k\pi$$
$$\frac{\pi}{6} + 2k\pi < x < \frac{11}{12}\pi  + 2k\pi$$
I'm strongly afraid it's wrong, but the fact is that the book provides other solutions I cannot managed to find. Unfortunately I don't remember them well, but I underwent the problem to Mathematica too, and it says that the system cannot be solved with the methods available to Reduce. Bah.

Hint: Going with the comment regarding use of the 'unit circle method', it may be easier to think about this geometrically rather than algebraically. For instance, if we pick a point on the upper half of the unit circle then the angle $x/2$ corresponds to a point in the first quadrant with positive sine. On the other hand, the condition that $\cos x>\sqrt{3}/2$ corresponds to a point on the unit circle which lies to the right of $x=\sqrt{3}/2$. By considering all the relevant cases, you should be able to determine the intervals in $[0,2\pi)$ for which $x$ satisfies the inequality.
