1: cot x=-6 nên cosx/sinx=-6
=>cosx=-6*sinx
\(F=\dfrac{sinx-3\cdot cosx}{cosx+2\cdot sinx}=\dfrac{sinx+18\cdot sinx}{-6\cdot sinx+2\cdot sinx}=\dfrac{20}{-4}=-5\)
2: cotx=1
=>cosx/sinx=1
=>cosx=sinx
\(I=\dfrac{sin^3x-4\cdot sin^3x}{sinx+3sinx}=\dfrac{5\cdot sin^3x}{4\cdot sinx}=\dfrac{5}{4}\cdot sin^2x\)
\(1+cot^2x=\dfrac{1}{sin^2x}\)
=>\(\dfrac{1}{sin^2x}=1+1=2\)
=>sin^2=1/2
=>\(I=\dfrac{5}{4}\cdot\dfrac{1}{2}=\dfrac{5}{8}\)
3: cotx=3
=>cosx/sinx=3
=>cosx=3*sinx
1+cot^2x=1/sin^2x
=>\(\dfrac{1}{sin^2x}=1+9=10\)
=>\(sin^2x=\dfrac{1}{10}\)
\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)
\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)
\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)
