Question

$2 sqrt{3} sin alpha -4cos alpha +6ctg ^{2} alpha$, если $sin alpha = frac{ sqrt{3} }{2}$, $0 leq alpha leq frac{ pi}{2}$

Answer 1

Если  $0 leq alpha leq frac{pi}{2}$  и  $sin alpha =frac{sqrt3}{2},$

то  $alpha =60^0$

$2sqrt3sin60^0-4cos60^0+6ctg^260^0=2sqrt3frac{sqrt3}{2}-4frac{1}{2}+6(frac{sqrt3}{3})^2}=\\=3-2+6cdot frac{1}{3}=3$