b) \(2sin^2x-3sinxcosx+cos^2x=0\)

\(\Leftrightarrow2tan^2x-3tanx+1=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{\pi}{4}\\tanx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\left(k\in Z\right)\)

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

=>\(\dfrac{1}{sin^2x}=1+9=10\)

=>\(sin^2x=\dfrac{1}{10}\)

\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)

\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)

\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)

Đặt \(\dfrac{x}{4}=t\)

\(2sin^22t-3cost=0\)

\(\Leftrightarrow8sin^2t.cos^2t-3cost=0\)

\(\Leftrightarrow8cos^2t\left(1-cos^2t\right)-3cost=0\)

\(\Leftrightarrow-8cos^4t+8cos^2t-3cost=0\)

\(\Leftrightarrow-cost\left(8cos^3t-8cost+3\right)=0\)

\(\Leftrightarrow cost\left(2cost-1\right)\left(4cos^2t+2cost-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\cost=\dfrac{1}{2}\\cost=\dfrac{-1+\sqrt{13}}{4}\\cost=\dfrac{-1-\sqrt{13}}{4}< -1\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow2cos2x.cos\left(\dfrac{\pi}{6}\right)-2sin2x.sin\left(\dfrac{\pi}{6}\right)+2sin2x-1=0\)

\(\Leftrightarrow\sqrt{3}cos2x+sin2x=1\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(tanx=3\) \(\Leftrightarrow sinx=3cosx\)

\(A=\dfrac{2.3.cosx-3cosx}{4cosx+5.3cosx}=\dfrac{3cosx}{19cosx}=\dfrac{3}{19}\)

\(B=\dfrac{sin^2x-4sinxcosx+3cos^2x}{5-2sin^2x}\)

\(=\dfrac{\left(3cosx\right)^2-4.3cosx.cosx+3cos^2x}{5-2\left(3cosx\right)^2}\)

\(=\dfrac{9cos^2x-12cos^2x+3cos^2x}{5-18cos^2x}=0\)

\(\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}=\dfrac{\dfrac{4sin\alpha}{cos\alpha}+\dfrac{5cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{4tan\alpha+5}{2tan\alpha-3}\)

Biết \(tan\)=\(\dfrac{1}{3}\) nên ta có:

\(\dfrac{4\times\dfrac{1}{2}+5}{2\times\dfrac{1}{2}-3}=\dfrac{2+5}{2-3}=\dfrac{7}{-2}=\dfrac{-7}{2}\)

\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)

\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)

Biết cotα=\(\dfrac{1}{2}\) nên ta có:

\(A=\dfrac{4+5\cdot\dfrac{1}{2}}{2-3\cdot\dfrac{1}{2}}\)

\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)

A= 13