Inequality with a linear sine-cosine factor and a tangent

Problem

Solve, studying the sign of each factor (on $[0,2\pi)$): $\frac{-\sin\alpha+\sqrt3\cos\alpha+\sqrt3}{-\tan\!\left(\dfrac{\alpha}{2}\right)-\sqrt3}<0.$

Solution

Numerator. $N=\sqrt3\cos\alpha-\sin\alpha+\sqrt3=2\cos\!\left(\alpha+\tfrac{\pi}{6}\right)+\sqrt3$. Then $N<0\Leftrightarrow\cos\!\left(\alpha+\tfrac{\pi}{6}\right)<-\tfrac{\sqrt3}{2}\Leftrightarrow\alpha\in\left(\tfrac{2\pi}{3},\pi\right)$; $N=0$ at $\alpha=\tfrac{2\pi}{3}$ and $\alpha=\pi$; elsewhere $N>0$. Denominator. $D=-\tan\tfrac{\alpha}{2}-\sqrt3<0\Leftrightarrow\tan\tfrac{\alpha}{2}>-\sqrt3$. On $[0,2\pi)$: $D<0$ on $[0,\pi)\cup\left(\tfrac{4\pi}{3},2\pi\right)$, $D>0$ on $\left(\pi,\tfrac{4\pi}{3}\right)$; $D=0$ at $\tfrac{4\pi}{3}$, undefined at $\pi$. Quotient $<0$ where $N$ and $D$ have opposite signs: on $[0,\tfrac{2\pi}{3})$ ($N>0,D<0$) and on $(\tfrac{4\pi}{3},2\pi)$ ($N>0,D<0$). $\boxed{\,0\le\alpha<\tfrac{2\pi}{3}\ \ \lor\ \ \tfrac{4\pi}{3}<\alpha<2\pi\quad(\text{mod }2\pi)\,}$