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.
 A: Taking the contrary approach of Semiclassical and following up on the comment of The Chaz 2.0, we can solve the inequality with a purely algebraic approach.
The product of 2 real numbers is negative if and only if both numbers are non zero and have different signs, so solving for
$$
\sin\left(\frac x 2\right) <0
\quad\&\quad
2\cos(x) -\sqrt 3 >0
$$
gives one set of solution, while
$$
\sin\left(\frac x 2\right) >0
\quad\&\quad
2\cos(x) -\sqrt 3 <0
$$
gives another.
As a non english native I'm not too sure what the unitary circle method is about so I'll refrain from commenting on that. But algebra does the trick.
A: HINT: use that $$\cos(x)=1-2\sin(x/2)^2$$ and you will get the inequality
$$\sin(x/2)(2-4\sin(x/2)^2-\sqrt{3})>0$$
A: 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.
A: $$\sin\left(\frac{x}{2}\right)\left(2\cos(x) - \sqrt{3}\right) >0 \iff\sin\left(\frac{x}{2}\right)\left(\cos(x) - \frac{\sqrt{3}}{2}\right) >0\\\iff\left(\sin\left(\frac{x}{2}\right)>0\text{ and }\cos(x) - \frac{\sqrt{3}}{2}>0\right)\text{ or } \left(\sin\left(\frac{x}{2}\right)<0\text{ and }\cos(x) - \frac{\sqrt{3}}{2}<0\right)\\\iff \left(\frac{x}{2}\in]0,\pi[+k2\pi\text{ and }x\in\left(\left]-\frac{\pi}{6},\frac{\pi}{6}\right[\cup\left]\frac{5\pi}{6},\frac{7\pi}{6}\right[\right)+k2\pi\right)\text{ or } \\\left(\frac{x}{2}\in]\pi,2\pi[+k2\pi\text{ and }x\in\left(\left]\frac{\pi}{6},\frac{5\pi}{6}\right[\cup\left]\frac{7\pi}{6},\frac{11\pi}{6}\right[\right)+k2\pi\right)\\
\iff\left(x\in\left(\left]0,\frac{\pi}{2}\right[\cup\left]\pi,\frac{3\pi}{2}\right[\right)+k2\pi\text{ and }x\in\left(\left]-\frac{\pi}{6},\frac{\pi}{6}\right[\cup\left]\frac{5\pi}{6},\frac{7\pi}{6}\right[\right)+k2\pi\right)\text{ or } \\\left(x\in\left(\left]\frac{\pi}{2},\pi\right[\cup\left]\frac{3\pi}{2},2\pi\right[\right)+k2\pi\text{ and }x\in\left(\left]\frac{\pi}{6},\frac{5\pi}{6}\right[\cup\left]\frac{7\pi}{6},\frac{11\pi}{6}\right[\right)+k2\pi\right)\\\iff x\in\left(\left]0,\frac{\pi}{6}\right[\cup\left]\pi,\frac{7\pi}{6}\right[\right)+k2\pi \text{ or } x\in\left(\left]\frac{\pi}{2},\frac{5\pi}{6}\right[\cup\left]\frac{3\pi}{2},\frac{11\pi}{6}\right[\right)+k2\pi\\\iff x\in\left(\left]0,\frac{\pi}{6}\right[\cup\left]\pi,\frac{7\pi}{6}\right[\cup\left]\frac{\pi}{2},\frac{5\pi}{6}\right[\cup\left]\frac{3\pi}{2},\frac{11\pi}{6}\right[\right)+k2\pi$$
A: Modulo $4\pi$: 


*

*either $\sin\dfrac x2>0\;$ and $\;\cos x>\dfrac{\sqrt 3}2=\cos\dfrac\pi6$, which means
$$0<x<2\pi\quad\text{and}\quad 0\le x<\frac\pi6\enspace\text{or}\enspace\frac{11\pi}6<x\le 2\pi$$

*or $\sin\dfrac x2<0\;$ and $\;\cos x<\dfrac{\sqrt 3}2$, which means
$$-2\pi<x<0\quad\text{and}\;-\frac{11\pi}6<x< -\frac{\pi}6.$$
Therefore the solutions are (modulo $4\pi$)
$$\Bigl(\frac{11\pi}6, -\frac{\pi}6\Bigr)\cup[0, \frac{\pi}6\Bigr)\cup\Bigl(\frac{11\pi}6,2\pi\Bigr].$$

