Не могу решить помогите)

$\frac{5\pi}{6}=5\cdot \frac{\pi}{6}=5\cdot 30^\circ =150^\circ \in 2\; chetvert'\\\\4cos\frac{5\pi}{6}sin\frac{5\pi}{6}+3tg^2\frac{5\pi}{6}=2sin(2\cdot \frac{5\pi}{6})+3tg(\pi-\frac{\pi}{6})=\\\\=2sin(2\pi-\frac{\pi}{3})+3(-tg\frac{\pi}{6})^2=-2sin\frac{\pi}{3}+3tg^2\frac{\pi}{6}=\\\\=-2\cdot \frac{\sqrt3}{2}+3\cdot (\frac{\sqrt3}{3})^2=-\sqrt3+1$

Ответ: С.