cos2x = 1- sin^x 
sin2x= 2sinxcosx 

Nhóm lại bình thường và giải thôi

1.

\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(2\left(1-t^2\right)+3t=0\)

\(\Leftrightarrow-2t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-\frac{1}{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{1}{2\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

2.

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(1-t^2-4t=4\)

\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

3.

\(\Leftrightarrow1+cosx+sinx+sinx.cosx=2\)

\(\Leftrightarrow2\left(sinx+cosx\right)+2sinx.cosx-2=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(2t+t^2-1-2=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow sinx+cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(sin^3x+2sinx+3cosx=0\)

\(\Leftrightarrow sin^3x-sinx+3sinx+3cosx=0\)

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

\(\Leftrightarrow-sinx.cosx+3\left(sinx+cosx\right)=0\)

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

\(\Leftrightarrow\left(sinx+cosx\right)^2-6\left(sinx+cosx\right)-1=0\)

\(\Leftrightarrow\left(sinx+cosx\right)^2-6\left(sinx+cosx\right)-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=3+\sqrt{10}\\sinx+cosx=3-\sqrt{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{3+\sqrt{10}}{\sqrt{2}}\left(l\right)\\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{3-\sqrt{10}}{\sqrt{2}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(\dfrac{3-\sqrt{10}}{\sqrt{2}}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(\dfrac{3-\sqrt{10}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(\dfrac{3-\sqrt{10}}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(\dfrac{3-\sqrt{10}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

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

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left|sin\left(x+\dfrac{\pi}{4}\right)\right|=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\pm\dfrac{\sqrt{3}}{2}\)

...

\(A=\frac{2sin2x-2sin2x.cos2x}{2sin2x+2sin2x.cos2x}=\frac{1-cos2x}{1+cos2x}=\frac{2sin^2x}{2cos^2x}=tan^2x\)

\(B=\frac{2cos4x.sinx}{2cos4x}=sinx\)

Câu C ko dịch được đề

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

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

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

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

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

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