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/

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

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

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

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

\(\frac{\left(sin3x+cosx\right)sin3x+\left(cos3x+sinx\right)cos3x}{cos4x}\)

\(=\frac{sin^23x+sin3x.cosx+cos^23x+cos3x.sinx}{cos4x}=\frac{1+sin3x.cosx+cos3x.sinx}{cos4x}\)

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

\(=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{1+\frac{sin2x}{cos2x}}{1-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)

a.

\(\Leftrightarrow cosx\left[1-\left(1-2sin^2x\right)\right]-sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx-sin^2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)

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

b.

Câu b chắc chắn đề đúng chứ bạn? Vế phải ấy?

c/

\(\left(1+cosx\right)\left(sinx-cosx+3\right)=1-cos^2x\)

\(\Leftrightarrow\left(1+cosx\right)\left(sinx-cosx+3\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

d.

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

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

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

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

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

ĐKXĐ: ...

\(sin3x-cos3x+sinx+cosx=\dfrac{sin3x-cos3x+sinx+cosx}{\left(sin3x+cosx\right)\left(cos3x-sinx\right)}\)

\(\Rightarrow\left[{}\begin{matrix}sin3x-cos3x+sinx+cosx=0\left(1\right)\\\left(sin3x+cosx\right)\left(cos3x-sinx\right)=1\left(2\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow3sinx-4sin^3x-4cos^3x+3cosx+sinx+cosx=0\)

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

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

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

\(\Leftrightarrow sinx+cosx=0\) (loại)

(2) \(\Leftrightarrow sin3x.cos3x-sinx.cosx-sin3x.sinx+cos3x.cosx=1\)

\(\Leftrightarrow\dfrac{1}{2}sin6x-\dfrac{1}{2}sin2x+cos4x=1\)

\(\Leftrightarrow\dfrac{1}{2}\left(3sin2x-4sin^32x\right)-\dfrac{1}{2}sin2x+1-2sin^22x=1\)

\(\Leftrightarrow sin2x-2sin^32x-2sin^22x=0\)

\(\Leftrightarrow-sin2x\left(2sin^22x+2sin2x-1\right)=0\)

\(\Leftrightarrow...\)

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

\(sin3x-cos3x=\left(3sinx-4sin^3x\right)-\left(4cos^3x-3cosx\right)\)

\(=3\left(sinx+cosx\right)-4\left(sin^3x+cos^3x\right)\)

\(=2\left(sin^3x+cos^3x\right)-6\left(sin^3x+cos^3x\right)+3\left(sinx+cosx\right)\)

\(=2\left(sin^3x+cos^3x\right)-6\left(sinx+cosx\right)\left(1-sinx.cosx\right)+3\left(sinx+cosx\right)\)

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

\(=2\left(sin^3x+cos^3x\right)+6sinx.cosx\left(sinx+cosx\right)-3\left(sinx+cosx\right)\)

\(=2\left(sinx+cosx\right)^3-3\left(sinx+cosx\right)\) (đpcm)