Question

Дано: ctg(x) =2

Найти: (sin^4(x)+5sin^3(x)×cos(x)-cos^4(x) )/2sin^2(x)-3cos^2(x)

Answer 1

$ctgx=2\\\\\frac{sin^4x+5sin^3x\, cosx-cos^4x}{2sin^2x-3cos^2x}=\frac{sin^4x\cdot (1+5\cdot \frac{cosx}{sinx}-\frac{cos^4x}{sin^4x})}{2(1-cos^2x)-3cos^2x}=\frac{sin^4x\cdot (1+5\, ctgx-ctg^4x)}{2-5\, cos^2x}=\\\\=\frac{sin^4x\cdot (1+5\cdot 2-2^4)}{sin^2x\cdot (\frac{2}{sin^2x}-5\cdot \frac{cos^2x}{sin^2x})}=\frac{sin^2x\cdot (-5)}{2\cdot (1+ctg^2x)-5\cdot ctg^2x}=\frac{\frac{1}{1+ctg^2x}\cdot (-5)}{2\cdot (1+4)-5\cdot 4}=\\\\\\=\frac{-\frac{5}{5}}{-10}=\frac{1}{10}=0,1$

Answer 2

Решение и ответ смотри во вложении