1,\(VT=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}+\dfrac{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)\(=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)^2+cos^2\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)

\(=\dfrac{1}{\dfrac{1}{2}.sin\left(\dfrac{\pi}{2}+x\right)}=\dfrac{2}{cosx}=VP\)

2,\(VT=\left(sin^4x-cos^4x\right)\left(sin^4x+cos^4x\right)=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(=\left(sin^2-cos^2x\right)\left(1-2sin^2x.cos^2x\right)\)\(=-cos2x\left(1-\dfrac{1}{2}sin^22x\right)\)\(=-\dfrac{cos2x\left(2-sin^22x\right)}{2}=-\dfrac{cos2x\left(1+cos^22x\right)}{2}\)

\(VP=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)=-\dfrac{7}{8}cos2x-\dfrac{1}{8}\left[4cos^32x-3cos2x\right]=-\dfrac{7}{8}.cos2x-\dfrac{1}{2}cos^32x+\dfrac{3}{8}cos2x\)

\(=-\dfrac{1}{2}cos2x-\dfrac{1}{2}cos^32x=\dfrac{-cos2x\left(1+cos^22x\right)}{2}\)

\(\Rightarrow VT=VP\)(đpcm)

3, \(VT=3-4\left(1-2sin^2x\right)+1-2sin^22x=8sin^2x-2sin^22x=8sin^2x-8.sin^2x.cos^2x=8sin^2x\left(1-cos^2x\right)=8sin^4x=VP\)

4,\(VP=\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{2}\right)+sin\left(3x+\dfrac{\pi}{6}\right)\right]-\dfrac{1}{2}\left[cos\left(3x-\dfrac{\pi}{3}\right)+cos\left(x+\pi\right)\right]\)

\(=\dfrac{1}{2}\left(cosx+sin3x.\dfrac{\sqrt{3}}{2}+\dfrac{cos3x}{2}\right)-\dfrac{1}{2}\left(\dfrac{cos3x}{2}+sin3x.\dfrac{\sqrt{3}}{2}-cosx\right)\)

\(=\dfrac{1}{2}.2cosx=cosx=VP\)

5, \(VP=4cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\dfrac{\sqrt{3}}{2}+\dfrac{cosx}{2}\right)^2\)\(=cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\sqrt{3}+cosx\right)^2\)

\(VT=2.cos\left(2x-\dfrac{\pi}{6}\right)+2.sin\left(2x-\dfrac{\pi}{6}\right).cos\left(2x-\dfrac{\pi}{6}\right)=2cos\left(2x-\dfrac{\pi}{6}\right)\left[1+sin\left(2x-\dfrac{\pi}{6}\right)\right]\)

\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(1+\dfrac{sin2x.\sqrt{3}}{2}-\dfrac{cos2x}{2}\right)\)\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x+cos^2x+sinx.cosx.\sqrt{3}-\dfrac{cos^2x-sin^2x}{2}\right)\)

\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.\dfrac{3}{2}+sinx.cosx.\sqrt{3}+\dfrac{cos^2x}{2}\right)\)\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.3+2sinx.cosx.\sqrt{3}+cos^2x\right)\)

\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sinx.\sqrt{3}+cosx\right)^2\)

\(\Rightarrow VT=VP\) (dpcm)

1. 

ĐKXĐ: \(x\ne k\pi\)

\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)

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

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

2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.

3.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

4.

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

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

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

1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)

\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)

\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)

\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )

b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)

\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{1+2cos^2\dfrac{x}{2}-1+cos\dfrac{x}{2}}\)

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)

\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)

\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )

c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)

\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)

\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)

\(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)

\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)

\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )

d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)

\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )

a.

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

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

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

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

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

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

c.

\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)

\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)

\(\Leftrightarrow...\)

.