[latex]2 \sqrt{3} sin \alpha -4cos \alpha +6ctg ^{2} \alpha [/latex], если [latex]sin \alpha = \frac{ \sqrt{3} }{2} [/latex], [latex]0 \leq \alpha \leq \frac{ \pi}{2} [/latex]

Если  [latex]0 \leq \alpha \leq \frac{\pi}{2}[/latex]  и  [latex]sin \alpha =\frac{\sqrt3}{2},[/latex] то  [latex] \alpha =60^0[/latex] [latex]2\sqrt3sin60^0-4cos60^0+6ctg^260^0=2\sqrt3\frac{\sqrt3}{2}-4\frac{1}{2}+6(\frac{\sqrt3}{3})^2}=\\\\=3-2+6\cdot \frac{1}{3}=3[/latex]