\(\Leftrightarrow1+2sinx.cosx+1-cosx=2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx\)

\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx-2\right)=0\)

\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2sinx-1\right)\left(\sqrt{3}sinx+2\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(cosx+\sqrt{3}sinx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

1: tan x=3 nên sin x/cosx=3

=>sin x=3*cosx

\(B=\dfrac{2\cdot sinx-3cosx}{sinx+cosx}=\dfrac{2\cdot3\cdot cosx-3cosx}{3cosx+cosx}\)

\(=\dfrac{2\cdot3-3}{3+1}=\dfrac{3}{4}\)

2: tan x=-1 nên sin x/cosx=-1

=>sinx=-cosx

\(I=\dfrac{4\cdot\left(-cosx\right)^3+\left(cosx\right)^3}{-cosx+3\cdot cosx}=\dfrac{-3\cdot cos^3x}{2cosx}=-\dfrac{3}{2}\cdot cos^2x\)

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+1=2\)

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

=>I=-3/2*1/2=-3/4

\(y=\frac{sinx-cosx}{2sinx+cosx+3}\)

\(\Leftrightarrow2y.sinx+y.cosx+3y=sinx-cosx\)

\(\Leftrightarrow\left(2y-1\right)sinx+\left(y+1\right)cosx=-3y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(2y-1\right)^2+\left(y+1\right)^2\ge9y^2\)

\(\Leftrightarrow4y^2+2y-2\le0\Rightarrow-1\le y\le\frac{1}{2}\)

\(\Rightarrow y_{min}=-1\) khi \(sinx=-1\)