hộ vs ae ơi

a/ \(sin3x=sin\left(2x+x\right)=sin2xcosx+cos2x.sinx\)

\(=2sinxcos^2x+\left(1-2sin^2x\right)sinx=2sinx\left(1-sin^2x\right)+sinx-2sin^3x\)

\(=3sinx-4sin^3x\)

b/

\(tan2x+\frac{1}{cos2x}=\frac{sin2x}{cos2x}+\frac{1}{cos2x}=\frac{sin2x+1}{cos2x}=\frac{2sinxcosx+sin^2x+cos^2x}{cos^2x-sin^2x}\)

\(=\frac{\left(sinx+cosx\right)^2}{\left(sinx+cosx\right)\left(cosx-sinx\right)}=\frac{sinx+cosx}{cosx-sinx}=\frac{\left(sinx+cosx\right)\left(cosx-sinx\right)}{\left(cos-sinx\right)^2}\)

\(=\frac{cos^2x-sin^2x}{cos^2x+sin^2x-2sinxcosx}=\frac{1-2sin^2x}{1-sin2x}\)

c/

\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{cos^2x-sin^2x}\)

\(=\frac{2sinxcosx+2sinxcosx}{cos2x}=\frac{4sinxcosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)

d/

\(\frac{sin2x}{1+cos2x}=\frac{2sinxcosx}{1+2cos^2x-1}=\frac{2sinxcosx}{2cos^2x}=\frac{sinx}{cosx}=tanx\)

e/

a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

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

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

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

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)

b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)

c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)

\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)

a/ \(cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{4}{5}\)

\(\Rightarrow tanx=-\frac{3}{4}\Rightarrow A=\frac{129}{20}\)

b/ \(B=\frac{5sinx+3cosx}{3cosx-2sinx}=\frac{\frac{5sinx}{sinx}+\frac{3cosx}{sinx}}{\frac{3cosx}{sinx}-\frac{2sinx}{sinx}}=\frac{5+3cotx}{3cotx-2}=\frac{5+9}{9-2}\)

c/ \(C=\frac{sinx.cosx\left(cotx-2tanx\right)}{sinx.cosx\left(5cotx+tanx\right)}=\frac{cos^2x-2sin^2x}{5cos^2x+sin^2x}=\frac{cos^2x-2\left(1-cos^2x\right)}{5cos^2x+1-cos^2x}=\frac{3cos^2x-2}{4cos^2x+1}=...\)

d/ Không dịch được đề, ko biết mẫu số bên trái nó đến đâu cả

\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)

\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)

\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)

\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)

\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)

\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)

\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)

\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)

    1 + sinx + cosx + sin2x + cos2x = 0

<=> sin^2x+ cos^2 x + ( sinx+cosx) + 2.sinx.cosx + ( cos^2 x - sin^2 x)=0

<=> 2 cos^2 x + 2sinx.cosx + sinx + cosx =0

<=> 2cosx ( cos x + sinx) + sinx + cosx = 0

<=> ( cosx + sinx ) (2 cos x + 1 ) = 0

<=> cosx + sinx = 0 hoặc 2cosx + 1 =0